Exercise

The most basic form of comparison is equality. Let’s briefly recap its syntax. The following statements all evaluate to TRUE (feel free to try them out in the console).

3 == (2 + 1)
"intermediate" != "r"
TRUE != FALSE
"Rchitect" != "rchitect"

Notice from the last expression that R is case sensitive: “R” is not equal to “r”. Keep this in mind when solving the exercises in this chapter!

Instructions

• In the editor on the right, write R code to see if TRUE equals FALSE.
• Likewise, check if -6 * 14 is not equal to 17 - 101.
• Next up: comparison of character strings. Ask R whether the strings “useR” and “user” are equal.
• Finally, find out what happens if you compare logicals to numerics: are TRUE and 1 equal?

# Comparison of logicals
TRUE == FALSE

## [1] FALSE

# Comparison of numerics
(-6 * 14) != (17 - 101)

## [1] FALSE

# Comparison of character strings
"useR" == "user"

## [1] FALSE

# Compare a logical with a numeric
TRUE == 1

## [1] TRUE

Awesome! Since TRUE coerces to 1 under the hood, TRUE == 1 evaluates to TRUE. Make sure not to mix up == (comparison) and = (assignment), == is what need to check the equality of R objects.

Exercise

Apart from equality operators, Filip also introduced the less than and greater than operators: < and >. You can also add an equal sign to express less than or equal to or greater than or equal to, respectively. Have a look at the following R expressions, that all evaluate to FALSE:

(1 + 2) > 4
"dog" < "Cats"
TRUE <= FALSE

Remember that for string comparison, R determines the greater than relationship based on alphabetical order. Also, keep in mind that TRUE corresponds to 1 in R, and FALSE coerces to 0 behind the scenes. Therefore, FALSE < TRUE is TRUE.

Instructions

Write R expressions to check whether:

• -6 * 5 + 2 is greater than or equal to -10 + 1.
• “raining” is less than or equal to “raining dogs”.
• TRUE is greater than FALSE.

# Comparison of numerics
(-6 * 5 + 2) >= (-10 + 1)

## [1] FALSE

# Comparison of character strings
"raining" <= "raining dogs"

## [1] TRUE

# Comparison of logicals
TRUE > FALSE

## [1] TRUE

Exercise

You are already aware that R is very good with vectors. Without having to change anything about the syntax, R’s relational operators also work on vectors.

Let’s go back to the example that was started in the video. You want to figure out whether your activity on social media platforms have paid off and decide to look at your results for LinkedIn and Facebook. The sample code in the editor initializes the vectors linkedin and facebook. Each of the vectors contains the number of profile views your LinkedIn and Facebook profiles had over the last seven days.

Instructions

Using relational operators, find a logical answer, i.e. TRUE or FALSE, for the following questions:

• On which days did the number of LinkedIn profile views exceed 15?
• When was your LinkedIn profile viewed only 5 times or fewer?
• When was your LinkedIn profile visited more often than your Facebook profile?

# The linkedin and facebook vectors have already been created for you
linkedin <- c(16, 9, 13, 5, 2, 17, 14)
facebook <- c(17, 7, 5, 16, 8, 13, 14)

# Popular days
linkedin > 15

## [1]  TRUE FALSE FALSE FALSE FALSE  TRUE FALSE

# Quiet days
linkedin <= 5

## [1] FALSE FALSE FALSE  TRUE  TRUE FALSE FALSE

# LinkedIn more popular than Facebook
linkedin > facebook

## [1] FALSE  TRUE  TRUE FALSE FALSE  TRUE FALSE

Exercise

R’s ability to deal with different data structures for comparisons does not stop at vectors. Matrices and relational operators also work together seamlessly!

Instead of in vectors (as in the previous exercise), the LinkedIn and Facebook data is now stored in a matrix called views. The first row contains the LinkedIn information; the second row the Facebook information. The original vectors facebook and linkedin are still available as well.

Instructions

Using the relational operators you’ve learned so far, try to discover the following:

• When were the views exactly equal to 13? Use the views matrix to return a logical matrix.
• For which days were the number of views less than or equal to 14? Again, have R return a logical matrix.

# The social data has been created for you
linkedin <- c(16, 9, 13, 5, 2, 17, 14)
facebook <- c(17, 7, 5, 16, 8, 13, 14)
views <- matrix(c(linkedin, facebook), nrow = 2, byrow = TRUE)

# When does views equal 13?
views == 13

##       [,1]  [,2]  [,3]  [,4]  [,5]  [,6]  [,7]
## [1,] FALSE FALSE  TRUE FALSE FALSE FALSE FALSE
## [2,] FALSE FALSE FALSE FALSE FALSE  TRUE FALSE

# When is views less than or equal to 14?
views <= 14

##       [,1] [,2] [,3]  [,4] [,5]  [,6] [,7]
## [1,] FALSE TRUE TRUE  TRUE TRUE FALSE TRUE
## [2,] FALSE TRUE TRUE FALSE TRUE  TRUE TRUE

Exercise

Before you work your way through the next exercises, have a look at the following R expressions. All of them will evaluate to TRUE:

TRUE & TRUE
FALSE | TRUE
5 <= 5 & 2 < 3
3 < 4 | 7 < 6

Watch out: 3 < x < 7 to check if x is between 3 and 7 will not work; you’ll need 3 < x & x < 7 for that.

In this exercise, you’ll be working with the last variable. This variable equals the last value of the linkedin vector that you’ve worked with previously. The linkedin vector represents the number of LinkedIn views your profile had in the last seven days, remember? Both the variables linkedin and last have already been defined in the editor.

Instructions

Write R expressions to solve the following questions concerning the variable last:

• Is last under 5 or above 10?
• Is last between 15 and 20, excluding 15 but including 20?

# The linkedin and last variable are already defined for you
linkedin <- c(16, 9, 13, 5, 2, 17, 14)
last <- tail(linkedin, 1)

# Is last under 5 or above 10?
last < 5 | last > 10

## [1] TRUE

# Is last between 15 (exclusive) and 20 (inclusive)?
last > 15 & last <= 20

## [1] FALSE

Exercise

Like relational operators, logical operators work perfectly fine with vectors and matrices.

Both the vectors linkedin and facebook are available again. Also a matrix - views - has been defined; its first and second row correspond to the linkedin and facebook vectors, respectively. Ready for some advanced queries to gain more insights into your social outreach?

Instructions

• When did LinkedIn views exceed 10 and did Facebook views fail to reach 10 for a particular day? Use the linkedin and facebook vectors.
• When were one or both of your LinkedIn and Facebook profiles visited at least 12 times?
• When is the views matrix equal to a number between 11 and 14, excluding 11 and including 14?

# The social data (linkedin, facebook, views) has been created for you
linkedin

## [1] 16  9 13  5  2 17 14

facebook

## [1] 17  7  5 16  8 13 14

views

##      [,1] [,2] [,3] [,4] [,5] [,6] [,7]
## [1,]   16    9   13    5    2   17   14
## [2,]   17    7    5   16    8   13   14

# linkedin exceeds 10 but facebook below 10
linkedin > 10 & facebook < 10

## [1] FALSE FALSE  TRUE FALSE FALSE FALSE FALSE

# When were one or both visited at least 12 times?
linkedin >= 12 | facebook >= 12

## [1]  TRUE FALSE  TRUE  TRUE FALSE  TRUE  TRUE

# When is views between 11 (exclusive) and 14 (inclusive)?
views > 11 & views <= 14

##       [,1]  [,2]  [,3]  [,4]  [,5]  [,6] [,7]
## [1,] FALSE FALSE  TRUE FALSE FALSE FALSE TRUE
## [2,] FALSE FALSE FALSE FALSE FALSE  TRUE TRUE

Possible Answers

• TRUE
• FALSE
• Running this piece of code would throw an error.

Exercise

With the things you’ve learned by now, you’re able to solve pretty cool problems.

Instead of recording the number of views for your own LinkedIn profile, suppose you conducted a survey inside the company you’re working for. You’ve asked every employee with a LinkedIn profile how many visits their profile has had over the past seven days. You stored the results in a data frame called li_df. This data frame is available in the workspace; type li_df in the console to check it out.

Instructions

• Select the entire second column, named day2, from the li_df data frame as a vector and assign it to second.
• Use second to create a logical vector, that contains TRUE if the corresponding number of views is strictly greater than 25 or strictly lower than 5 and FALSE otherwise. Store this logical vector as extremes.
• Use sum() on the extremes vector to calculate the number of TRUEs in extremes (i.e. to calculate the number of employees that are either very popular or very low-profile). Simply print this number to the console.

# li_df is pre-loaded in your workspace
li_df <- data.frame(day1 = c(2, 19, 24, 22, 25, 22, 0, 12, 19, 23, 29, 13, 7, 26, 7, 32, 7, 9, 0, 9, 6, 17, 1, 5, 2, 29, 17, 26, 27, 4, 22, 9, 6, 18, 2, 32, 5, 6, 30, 34, 15, 28, 6, 17, 6, 18, 21, 10, 6, 30), day2 = c(3, 23, 18, 18, 25, 20, 4, 3, 22, 12, 27, 13, 17, 27, 6, 35, 17, 6, 1, 12, 15, 17, 12, 8, 7, 25, 15, 32, 29, 1, 22, 11, 5, 17, 12, 26, 13, 10, 37, 33, 19, 29, 8, 22, 10, 19, 27, 18, 15, 28), day3 = c(3, 18, 15, 27, 26, 29, 2, 15, 22, 19, 23, 20, 9, 28, 4, 31, 9, 3, 11, 6, 15, 12, 8, 0, 5, 32, 17, 33, 24, 1, 17, 7, 12, 12, 13, 20, 12, 11, 32, 32, 21, 30, 6, 27, 17, 22, 28, 20, 15, 29), day4 = c(6, 22, 19, 26, 31, 26, 2, 7, 19, 25, 25, 17, 5, 36, 11, 35, 12, 12, 6, 13, 10, 4, 2, 1, 3, 28, 23, 30, 29, 2, 20, 10, 5, 22, 7, 23, 11, 6, 35, 35, 18, 19, 7, 24, 18, 17, 28, 18, 15, 31), day5 = c(4, 23, 18, 19, 24, 23, 3, 1, 25, 18, 29, 12, 11, 29, 5, 24, 13, 3, 0, 12, 9, 14, 4, 6, 1, 28, 23, 33, 26, 1, 14, 8, 17, 22, 10, 24, 6, 6, 37, 33, 22, 21, 17, 18, 13, 21, 26, 12, 10, 24), day6 = c(2, 29, 22, 21, 36, 22, 4, 15, 24, 22, 30, 22, 9, 31, 5, 25, 6, 8, 4, 13, 7, 17, 4, 3, 5, 27, 17, 28, 31, 7, 19, 15, 17, 13, 6, 25, 5, 2, 41, 27, 26, 19, 11, 28, 10, 15, 17, 19, 14, 20), day7 = c(0, 25, 17, 25, 37, 29, 2, 11, 23, 22, 17, 20, 9, 30, 15, 36, 12, 6, 11, 11, 18, 7, 11, 1, 5, 27, 22, 26, 28, 4, 13, 5, 4, 12, 2, 21, 10, 5, 42, 35, 22, 26, 14, 24, 7, 23, 25, 17, 2, 25), row.names = c("employee_1", "employee_2", "employee_3", "employee_4", "employee_5", "employee_6", "employee_7", "employee_8", "employee_9", "employee_10", "employee_11", "employee_12", "employee_13", "employee_14", "employee_15", "employee_16", "employee_17", "employee_18", "employee_19", "employee_20", "employee_21", "employee_22", "employee_23", "employee_24", "employee_25", "employee_26", "employee_27", "employee_28", "employee_29", "employee_30", "employee_31", "employee_32", "employee_33", "employee_34", "employee_35", "employee_36", "employee_37", "employee_38", "employee_39", "employee_40", "employee_41", "employee_42", "employee_43", "employee_44", "employee_45", "employee_46", "employee_47", "employee_48", "employee_49", "employee_50"))

# Select the second column, named day2, from li_df: second
second <- li_df$day2

# Build a logical vector, TRUE if value in second is extreme: extremes
extremes <- (second > 25 | second < 5)

# Count the number of TRUEs in extremes
sum(extremes)

## [1] 16

# Solve it with a one-liner
sum(li_df$day2 > 25 | li_df$day2 < 5)

## [1] 16

Exercise

Before diving into some exercises on the if statement, have another look at its syntax:

if (condition) {
  expr
}

Remember your vectors with social profile views? Let’s look at it from another angle. The medium variable gives information about the social website; the num_views variable denotes the actual number of views that particular medium had on the last day of your recordings. Both these variables have already been defined in the editor.

Instructions

• Examine the if statement that prints out “Showing LinkedIn information” if the medium variable equals “LinkedIn”.
• Code an if statement that prints “You’re popular!” to the console if the num_views variable exceeds 15.

# Variables related to your last day of recordings
medium <- "LinkedIn"
num_views <- 14

# Examine the if statement for medium
if (medium == "LinkedIn") {
  print("Showing LinkedIn information")
}

## [1] "Showing LinkedIn information"

# Write the if statement for num_views
if (num_views > 15) {
  print("You're popular!")
}

Exercise

You can only use an else statement in combination with an if statement. The else statement does not require a condition; its corresponding code is simply run if all of the preceding conditions in the control structure are FALSE. Here’s a recipe for its usage:

if (condition) {
  expr1
} else {
  expr2
}

It’s important that the else keyword comes on the same line as the closing bracket of the if part!

Both if statements that you coded in the previous exercises are already available in the editor. It’s now up to you to extend them with the appropriate else statements!

Instructions

Add an else statement to both control structures, such that

• “Unknown medium” gets printed out to the console when the if-condition on medium does not hold.
• R prints out “Try to be more visible!” when the if-condition on num_views is not met.

# Variables related to your last day of recordings
medium <- "LinkedIn"
num_views <- 14

# Control structure for medium
if (medium == "LinkedIn") {
  print("Showing LinkedIn information")
} else {
  print("Unknown medium")
}

## [1] "Showing LinkedIn information"

# Control structure for num_views
if (num_views > 15) {
  print("You're popular!")
} else {
  print("Try to be more visible!")
}

## [1] "Try to be more visible!"

Exercise

The else if statement allows you to further customize your control structure. You can add as many else if statements as you like. Keep in mind that R ignores the remainder of the control structure once a condition has been found that is TRUE and the corresponding expressions have been executed. Here’s an overview of the syntax to freshen your memory:

if (condition1) {
  expr1
} else if (condition2) {
  expr2
} else if (condition3) {
  expr3
} else {
  expr4
}

Again, It’s important that the else if keywords comes on the same line as the closing bracket of the previous part of the control construct!

Instructions

Add code to both control structures such that:

• R prints out “Showing Facebook information” if medium is equal to “Facebook”. Remember that R is case sensitive!
• “Your number of views is average” is printed if num_views is between 15 (inclusive) and 10 (exclusive). Feel free to change the variables medium and num_views to see how the control structure respond. In both cases, the existing code should be extended in the else if statement. No existing code should be modified.

# Variables related to your last day of recordings
medium <- "LinkedIn"
num_views <- 14

# Control structure for medium
if (medium == "LinkedIn") {
  print("Showing LinkedIn information")
} else if (medium == "Facebook") {
  # Add code to print correct string when condition is TRUE
  print("Showing Facebook information")
} else {
  print("Unknown medium")
}

## [1] "Showing LinkedIn information"

# Control structure for num_views
if (num_views > 15) {
  print("You're popular!")
} else if (num_views <= 15 & num_views > 10) {
  # Add code to print correct string when condition is TRUE
  print("Your number of views is average")
} else {
  print("Try to be more visible!")
}

## [1] "Your number of views is average"

Awesome! Have another look at the second control structure. Because R abandons the control flow as soon as it finds a condition that is met, you can simplify the condition for the else if part in the second construct to num_views > 10.

Possible Answers

• 2 and 4
• 1 and 4
• 1 and 3
• 2 and 3

Exercise

In this exercise, you will combine everything that you’ve learned so far: relational operators, logical operators and control constructs. You’ll need it all!

In the editor, we’ve coded two values beforehand: li and fb, denoting the number of profile views your LinkedIn and Facebook profile had on the last day of recordings. Go through the instructions to create R code that generates a ‘social media score’, sms, based on the values of li and fb.

Instructions

Finish the control-flow construct with the following behavior:

• If both li and fb are 15 or higher, set sms equal to double the sum of li and fb.
• If both li and fb are strictly below 10, set sms equal to half the sum of li and fb.
• In all other cases, set sms equal to li + fb.
• Finally, print the resulting sms variable to the console.

# Variables related to your last day of recordings
li <- 15
fb <- 9

# Code the control-flow construct
if (li >= 15 & fb >= 15) {
  sms <- 2 * (li + fb)
} else if (li < 10 & fb < 10) {
  sms <- 0.5 * (li + fb)
} else {
  sms <- sum(li, fb)
}

# Print the resulting sms to the console
sms

## [1] 24

Exercise

Let’s get you started with building a while loop from the ground up. Have another look at its recipe:

while (condition) {
  expr
}

Remember that the condition part of this recipe should become FALSE at some point during the execution. Otherwise, the while loop will go on indefinitely. In DataCamp’s learning interface, your session will be disconnected in this case.

Have a look at the code on the right; it initializes the speed variables and already provides a while loop template to get you started.

Instructions

Code a while loop with the following characteristics:

• The condition of the while loop should check if speed is higher than 30.
• Inside the body of the while loop, print out "Slow down!".
• Inside the body of the while loop, decrease the speed by 7 units. This step is crucial; otherwise your while loop will never stop.

# Initialize the speed variable
speed <- 64

# Code the while loop
while (speed > 30) {
  print("Slow down!")
  speed = speed -7
}

## [1] "Slow down!"
## [1] "Slow down!"
## [1] "Slow down!"
## [1] "Slow down!"
## [1] "Slow down!"

# Print out the speed variable
speed

## [1] 29

Exercise

In the previous exercise, you simulated the interaction between a driver and a driver’s assistant: When the speed was too high, “Slow down!” got printed out to the console, resulting in a decrease of your speed by 7 units.

There are several ways in which you could make your driver’s assistant more advanced. For example, the assistant could give you different messages based on your speed or provide you with a current speed at a given moment.

A while loop similar to the one you’ve coded in the previous exercise is already available in the editor. It prints out your current speed, but there’s no code that decreases the speed variable yet, which is pretty dangerous. Can you make the appropriate changes?

Instructions

• If the speed is greater than 48, have R print out “Slow down big time!”, and decrease the speed by 11.
• Otherwise, have R simply print out “Slow down!”, and decrease the speed by 6.

# Initialize the speed variable
speed <- 64

# Extend/adapt the while loop
while (speed > 30) {
  print(paste("Your speed is",speed))
  if (speed > 48) {
    print("Slow down big time!")
    speed = speed - 11
  } else {
    print("Slow down!")
    speed = speed - 6
  }
}

## [1] "Your speed is 64"
## [1] "Slow down big time!"
## [1] "Your speed is 53"
## [1] "Slow down big time!"
## [1] "Your speed is 42"
## [1] "Slow down!"
## [1] "Your speed is 36"
## [1] "Slow down!"

Exercise

There are some very rare situations in which severe speeding is necessary: what if a hurricane is approaching and you have to get away as quickly as possible? You don’t want the driver’s assistant sending you speeding notifications in that scenario, right?

This seems like a great opportunity to include the break statement in the while loop you’ve been working on. Remember that the break statement is a control statement. When R encounters it, the while loop is abandoned completely.

Instructions

Adapt the while loop such that it is abandoned when the speed of the vehicle is greater than 80. This time, the speed variable has been initialized to 88; keep it that way.

# Initialize the speed variable
speed <- 88

while (speed > 30) {
  print(paste("Your speed is", speed))
  
  # Break the while loop when speed exceeds 80
  if (speed > 80) {
    break
  }
  
  if (speed > 48) {
    print("Slow down big time!")
    speed <- speed - 11
  } else {
    print("Slow down!")
    speed <- speed - 6
  }
}

## [1] "Your speed is 88"

Exercise

The previous exercises guided you through developing a pretty advanced while loop, containing a break statement and different messages and updates as determined by control flow constructs. If you manage to solve this comprehensive exercise using a while loop, you’re totally ready for the next topic: the for loop.

Instructions

Finish the while loop so that it:

• prints out the triple of i, so 3 * i, at each run.
• is abandoned with a break if the triple of i is divisible by 8, but still prints out this triple before breaking.

# Initialize i as 1 
i <- 1

# Code the while loop
while (i <= 10) {
  print(3 * i)
  if ((3 * i) %% 8 == 0) {
    print(3 * i)
    break
  }
  i <- i + 1
}

## [1] 3
## [1] 6
## [1] 9
## [1] 12
## [1] 15
## [1] 18
## [1] 21
## [1] 24
## [1] 24

Exercise

In the previous video, Filip told you about two different strategies for using the for loop. To refresh your memory, consider the following loops that are equivalent in R:

primes <- c(2, 3, 5, 7, 11, 13)

# loop version 1
for (p in primes) {
  print(p)
}

# loop version 2
for (i in 1:length(primes)) {
  print(primes[i])
}

Remember our linkedin vector? It’s a vector that contains the number of views your LinkedIn profile had in the last seven days. The linkedin vector has already been defined in the editor on the right so that you can fully focus on the instructions!

Instructions

Write a for loop that iterates over all the elements of linkedin and prints out every element separately. Do this in two ways: using the loop version 1 and the loop version 2 in the example code above.

# The linkedin vector has already been defined for you
linkedin <- c(16, 9, 13, 5, 2, 17, 14)

# Loop version 1
for (i in linkedin) {
  print(i)
}

## [1] 16
## [1] 9
## [1] 13
## [1] 5
## [1] 2
## [1] 17
## [1] 14

# Loop version 2
for (i in 1:length(linkedin)) {
  print(linkedin[i])
}

## [1] 16
## [1] 9
## [1] 13
## [1] 5
## [1] 2
## [1] 17
## [1] 14

Exercise

Looping over a list is just as easy and convenient as looping over a vector. There are again two different approaches here:

primes_list <- list(2, 3, 5, 7, 11, 13)

# loop version 1
for (p in primes_list) {
  print(p)
}

# loop version 2
for (i in 1:length(primes_list)) {
  print(primes_list[[i]])
}

Notice that you need double square brackets - [[ ]] - to select the list elements in loop version 2.

Suppose you have a list of all sorts of information on New York City: its population size, the names of the boroughs, and whether it is the capital of the United States. We’ve already prepared a list nyc with all this information in the editor (source: Wikipedia).

Instructions

As in the previous exercise, loop over the nyc list in two different ways to print its elements:

• Loop directly over the nyc list (loop version 1).
• Define a looping index and do subsetting using double brackets (loop version 2).

# The nyc list is already specified
nyc <- list(pop = 8405837, 
            boroughs = c("Manhattan", "Bronx", "Brooklyn", "Queens", "Staten Island"), 
            capital = FALSE)

# Loop version 1
for (i in nyc) {
  print(i)
}

## [1] 8405837
## [1] "Manhattan"     "Bronx"         "Brooklyn"      "Queens"       
## [5] "Staten Island"
## [1] FALSE

# Loop version 2
for (i in 1:length(nyc)) {
  print(nyc[[i]])
}

## [1] 8405837
## [1] "Manhattan"     "Bronx"         "Brooklyn"      "Queens"       
## [5] "Staten Island"
## [1] FALSE

Exercise

In your workspace, there’s a matrix ttt, that represents the status of a tic-tac-toe game. It contains the values “X”, “O” and “NA”. Print out ttt in the console so you can have a closer look. On row 1 and column 1, there’s “O”, while on row 3 and column 2 there’s “NA”.

To solve this exercise, you’ll need a for loop inside a for loop, often called a nested loop. Doing this in R is a breeze! Simply use the following recipe:

for (var1 in seq1) {
  for (var2 in seq2) {
    expr
  }
}

Instructions

Finish the nested for loops to go over the elements in ttt:

• The outer loop should loop over the rows, with loop index i (use 1:nrow(ttt)).
• The inner loop should loop over the columns, with loop index j (use 1:ncol(ttt)).
• Inside the inner loop, make use of print() and paste() to print out information in the following format: “On row i and column j the board contains x”, where x is the value on that position.

# The tic-tac-toe matrix ttt has already been defined for you
ttt <- matrix(c("0", NA, "X", NA, "0", "0", "X", NA, "X"), nrow=3, byrow=TRUE)

# define the double for loop
for (i in 1:nrow(ttt)) {
  for (j in 1:ncol(ttt)) {
    print(paste("On row ", i, " and column ", j, " the board contains ", ttt[i,j]))
  }
}

## [1] "On row  1  and column  1  the board contains  0"
## [1] "On row  1  and column  2  the board contains  NA"
## [1] "On row  1  and column  3  the board contains  X"
## [1] "On row  2  and column  1  the board contains  NA"
## [1] "On row  2  and column  2  the board contains  0"
## [1] "On row  2  and column  3  the board contains  0"
## [1] "On row  3  and column  1  the board contains  X"
## [1] "On row  3  and column  2  the board contains  NA"
## [1] "On row  3  and column  3  the board contains  X"

Exercise

Let’s return to the LinkedIn profile views data, stored in a vector linkedin. In the first exercise on for loops you already did a simple printout of each element in this vector. A little more in-depth interpretation of this data wouldn’t hurt, right? Time to throw in some conditionals! As with the while loop, you can use the if and else statements inside the for loop.

Instructions

Add code to the for loop that loops over the elements of the linkedin vector:

• If the vector element’s value exceeds 10, print out “You’re popular!”.
• If the vector element’s value does not exceed 10, print out “Be more visible!”

# The linkedin vector has already been defined for you
linkedin <- c(16, 9, 13, 5, 2, 17, 14)

# Code the for loop with conditionals
for (li in linkedin) {
  if (li > 10) {
    print("You're popular!")
  } else {
    print("Be more visible!")
  }
  print(li)
}

## [1] "You're popular!"
## [1] 16
## [1] "Be more visible!"
## [1] 9
## [1] "You're popular!"
## [1] 13
## [1] "Be more visible!"
## [1] 5
## [1] "Be more visible!"
## [1] 2
## [1] "You're popular!"
## [1] 17
## [1] "You're popular!"
## [1] 14

Exercise

In the editor on the right you’ll find a possible solution to the previous exercise. The code loops over the linkedin vector and prints out different messages depending on the values of li.

In this exercise, you will use the break and next statements:

• The break statement abandons the active loop: the remaining code in the loop is skipped and the loop is not iterated over anymore.
• The next statement skips the remainder of the code in the loop, but continues the iteration.

Instructions

Extend the for loop with two new, separate if tests in the editor as follows:

• If the vector element’s value exceeds 16, print out “This is ridiculous, I’m outta here!” and have R abandon the for loop (break).
• If the value is lower than 5, print out “This is too embarrassing!” and fast-forward to the next iteration (next).

# The linkedin vector has already been defined for you
linkedin <- c(16, 9, 13, 5, 2, 17, 14)

# Extend the for loop
for (li in linkedin) {
  if (li > 10) {
    print("You're popular!")
  } else {
    print("Be more visible!")
  }
  
  # Add if statement with break
  if (li > 16) {
    print("This is ridiculous! I'm outta here!")
    break
  }
  
  # Add if statement with next
  if (li < 5) {
    print("This is too embarrassing!")
    next
  }
  
  print(li)
}

## [1] "You're popular!"
## [1] 16
## [1] "Be more visible!"
## [1] 9
## [1] "You're popular!"
## [1] 13
## [1] "Be more visible!"
## [1] 5
## [1] "Be more visible!"
## [1] "This is too embarrassing!"
## [1] "You're popular!"
## [1] "This is ridiculous! I'm outta here!"

Exercise

This exercise will not introduce any new concepts on for loops.

In the editor on the right, we already went ahead and defined a variable rquote. This variable has been split up into a vector that contains separate letters and has been stored in a vector chars with the strsplit() function.

Can you write code that counts the number of r’s that come before the first u in rquote?

Instructions

• Initialize the variable rcount, as 0.
• Finish the for loop:
• if char equals "r", increase the value of rcount by 1.
• if char equals "u", leave the for loop entirely with a break.
• Finally, print out the variable rcount to the console to see if your code is correct.

# Pre-defined variables
rquote <- "r's internals are irrefutably intriguing"
chars <- strsplit(rquote, split = "")[[1]]

# Initialize rcount
rcount <- 0

# Finish the for loop
for (char in chars) {
  if (char == "r") {
    rcount = rcount + 1
  } else if (char == "u") {
    break
  }
}

# Print out rcount
rcount

## [1] 5

Exercise

Before even thinking of using an R function, you should clarify which arguments it expects. All the relevant details such as a description, usage, and arguments can be found in the documentation. To consult the documentation on the sample() function, for example, you can use one of following R commands:

help(sample)
?sample

If you execute these commands in the console of the DataCamp interface, you’ll be redirected to www.rdocumentation.org.

A quick hack to see the arguments of the sample() function is the args() function. Try it out in the console:

args(sample)

In the next exercises, you’ll be learning how to use the mean() function with increasing complexity. The first thing you’ll have to do is get acquainted with the mean() function.

Instructions

• Consult the documentation on the mean() function: ?mean or help(mean).
• Inspect the arguments of the mean() function using the args() function.

# Consult the documentation on the mean() function
?mean

## starting httpd help server ... done

# Inspect the arguments of the mean() function
args(mean)

## function (x, ...) 
## NULL

Exercise

The documentation on the mean() function gives us quite some information:

• The mean() function computes the arithmetic mean.
• The most general method takes multiple arguments: x and ....
• The x argument should be a vector containing numeric, logical or time-related information.

Remember that R can match arguments both by position and by name. Can you still remember the difference? You’ll find out in this exercise!

Once more, you’ll be working with the view counts of your social network profiles for the past 7 days. These are stored in the linkedin and facebook vectors and have already been defined in the editor on the right.

Instructions

• Calculate the average number of views for both linkedin and facebook and assign the result to avg_li and avg_fb, respectively. Experiment with different types of argument matching!
• Print out both avg_li and avg_fb.

# The linkedin and facebook vectors have already been created for you
linkedin <- c(16, 9, 13, 5, 2, 17, 14)
facebook <- c(17, 7, 5, 16, 8, 13, 14)

# Calculate average number of views
avg_li <- mean(linkedin)
avg_fb <- mean(facebook)

# Inspect avg_li and avg_fb
avg_li

## [1] 10.85714

avg_fb

## [1] 11.42857

Exercise

Check the documentation on the mean() function again:

?mean

The Usage section of the documentation includes two versions of the mean() function. The first usage,

mean(x, ...)

is the most general usage of the mean function. The ‘Default S3 method’, however, is:

mean(x, trim = 0, na.rm = FALSE, ...)

The ... is called the ellipsis. It is a way for R to pass arguments along without the function having to name them explicitly. The ellipsis will be treated in more detail in future courses.

For the remainder of this exercise, just work with the second usage of the mean function. Notice that both trim and na.rm have default values. This makes them optional arguments.

Instructions

• Calculate the mean of the element-wise sum of linkedin and facebook and store the result in a variable avg_sum.
• Calculate the mean once more, but this time set the trim argument equal to 0.2 and assign the result to avg_sum_trimmed.
• Print out both avg_sum and avg_sum_trimmed; can you spot the difference?

# The linkedin and facebook vectors have already been created for you
linkedin <- c(16, 9, 13, 5, 2, 17, 14)
facebook <- c(17, 7, 5, 16, 8, 13, 14)

# Calculate the mean of the sum
avg_sum <- mean(linkedin + facebook)

# Calculate the trimmed mean of the sum
avg_sum_trimmed <- mean(linkedin + facebook, trim = 0.2)

# Inspect both new variables
avg_sum

## [1] 22.28571

avg_sum_trimmed

## [1] 22.6

Exercise

In the video, Filip guided you through the example of specifying arguments of the sd() function. The sd() function has an optional argument, na.rm that specified whether or not to remove missing values from the input vector before calculating the standard deviation.

If you’ve had a good look at the documentation, you’ll know by now that the mean() function also has this argument, na.rm, and it does the exact same thing. By default, it is set to FALSE, as the Usage of the default S3 method shows:

mean(x, trim = 0, na.rm = FALSE, ...)

Let’s see what happens if your vectors linkedin and facebook contain missing values (NA).

Instructions

• Calculate the average number of LinkedIn profile views, without specifying any optional arguments. Simply print the result to the console.
• Calculate the average number of LinkedIn profile views, but this time tell R to strip missing values from the input vector.

# The linkedin and facebook vectors have already been created for you
linkedin <- c(16, 9, 13, 5, NA, 17, 14)
facebook <- c(17, NA, 5, 16, 8, 13, 14)

# Basic average of linkedin
mean(linkedin)

## [1] NA

# Advanced average of linkedin
mean(linkedin, na.rm = TRUE)

## [1] 12.33333

Exercise

You already know that R functions return objects that you can then use somewhere else. This makes it easy to use functions inside functions, as you’ve seen before:

speed <- 31
print(paste("Your speed is", speed))

Notice that both the print() and paste() functions use the ellipsis - ... - as an argument. Can you figure out how they’re used?

Instructions

Use abs() on linkedin - facebook to get the absolute differences between the daily Linkedin and Facebook profile views. Next, use this function call inside mean() to calculate the Mean Absolute Deviation. In the mean() call, make sure to specify na.rm to treat missing values correctly!

# The linkedin and facebook vectors have already been created for you
linkedin <- c(16, 9, 13, 5, NA, 17, 14)
facebook <- c(17, NA, 5, 16, 8, 13, 14)

# Calculate the mean absolute deviation
abs(linkedin - facebook)

## [1]  1 NA  8 11 NA  4  0

mean(abs(linkedin - facebook), na.rm = TRUE)

## [1] 4.8

Possible Answers

• (1) and (3)
• 2. and (4)
• (1), (2), and (3)
• (1), (2), and (4)

Exercise

Wow, things are getting serious… you’re about to write your own function! Before you have a go at it, have a look at the following function template:

my_fun <- function(arg1, arg2) {
  body
}

Notice that this recipe uses the assignment operator (<-) just as if you were assigning a vector to a variable for example. This is not a coincidence. Creating a function in R basically is the assignment of a function object to a variable! In the recipe above, you’re creating a new R variable my_fun, that becomes available in the workspace as soon as you execute the definition. From then on, you can use the my_fun as a function.

Instructions

• Create a function pow_two(): it takes one argument and returns that number squared (that number times itself).
• Call this newly defined function with 12 as input.
• Next, create a function sum_abs(), that takes two arguments and returns the sum of the absolute values of both arguments.
• Finally, call the function sum_abs() with arguments -2 and 3 afterwards.

# Create a function pow_two()
pow_two <- function(a) {
  a^2
}

# Use the function
pow_two(12)

## [1] 144

# Create a function sum_abs()
sum_abs <- function(x, y) {
  sum(abs(x),abs(y))
}

# Use the function
sum_abs(-2, 3)

## [1] 5

Exercise

There are situations in which your function does not require an input. Let’s say you want to write a function that gives us the random outcome of throwing a fair die:

throw_die <- function() {
  number <- sample(1:6, size = 1)
  number
}

throw_die()

Up to you to code a function that doesn’t take any arguments!

Instructions

• Define a function, hello(). It prints out “Hi there!” and returns TRUE. It has no arguments.
• Call the function hello(), without specifying arguments of course.

# Define the function hello()
hello <- function() {
  print("Hi there!")
  TRUE
}

# Call the function hello()
hello()

## [1] "Hi there!"

## [1] TRUE

Exercise

Do you still remember the difference between an argument with and without default values? Have another look at the sd() function by typing ?sd in the console. The usage section shows the following information:

sd(x, na.rm = FALSE)

This tells us that x has to be defined for the sd() function to be called correctly, however, na.rm already has a default value. Not specifying this argument won’t cause an error.

You can define default argument values in your own R functions as well. You can use the following recipe to do so:

my_fun <- function(arg1, arg2 = val2) {
  body
}

The editor on the right already includes an extended version of the pow_two() function from before. Can you finish it?

Instructions

• Add an optional argument, named print_info, that is TRUE by default.
• Wrap an if construct around the print() function: this function should only be executed if print_info is TRUE.
• Feel free to experiment with the pow_two() function you’ve just coded.

# Finish the pow_two() function
pow_two <- function(x, print_info = TRUE) {
  y <- x ^ 2
  if (print_info == TRUE) {
      print(paste(x, "to the power two equals", y))
      return(y)
  } else {
    return(y)
  }
}

pow_two(2, print_info = TRUE)

## [1] "2 to the power two equals 4"

## [1] 4

Possible Answers

• Executing two_dice() causes an error.
• Executing res <- two_dice() makes the contents of dice1 and dice2 available outside the function.
• Whatever the way of calling the two_dice() function, R won’t have access to dice1 and dice2 outside the function.

Possible Answers

• a and b equal 7 and 6 respectively after executing this code block.
• After the first call of increment(), where a is defined, a equals 7 and count equals 5.
• In the end, count will equal 10.
• In the last expression, the value of count was actually changed because of the explicit assignment.

Well done! Given that R passes arguments by value and not by reference, the value of count is not changed after the first two calls of increment(). Only in the final expression, where count is re-assigned explicitly, does the value of count change.

Exercise

Now that you’ve acquired some skills in defining functions with different types of arguments and return values, you should try to create more advanced functions. As you’ve noticed in the previous exercises, it’s perfectly possible to add control-flow constructs, loops and even other functions to your function body.

Remember our social media example? The vectors linkedin and facebook are already defined in the workspace so you can get your hands dirty straight away. As a first step, you will be writing a function that can interpret a single value of this vector. In the next exercise, you will write another function that can handle an entire vector at once.

Instructions

• Finish the function definition for interpret(), that interprets the number of profile views on a single day:
• The function takes one argument, num_views.
• If num_views is greater than 15, the function prints out “You’re popular!” to the console and returns num_views.
• Else, the function prints out “Try to be more visible!” and returns 0.
• Finally, call the interpret() function twice: on the first value of the linkedin vector and on the second element of the facebook vector.

# The linkedin and facebook vectors have already been created for you
linkedin <- c(16, 9, 13, 5, 2, 17, 14)
facebook <- c(17, 7, 5, 16, 8, 13, 14)

# Define the interpret function
interpret <- function(num_views) {
  if (num_views > 15) {
    print("You're popular!")
    return(num_views)
  } else {
    print("Try to be more visible!")
    return(0)
  }
}

# Call the interpret function twice
interpret(linkedin[1])

## [1] "You're popular!"

## [1] 16

interpret(facebook[2])

## [1] "Try to be more visible!"

## [1] 0

Funkadelic! The annoying thing here is that interpret() only takes one argument. Proceed to the next exercise to implement something more useful.

Exercise

A possible implementation of the interpret() function is already available in the editor. In this exercise you’ll be writing another function that will use the interpret() function to interpret all the data from your daily profile views inside a vector. Furthermore, your function will return the sum of views on popular days, if asked for. A for loop is ideal for iterating over all the vector elements. The ability to return the sum of views on popular days is something you can code through a function argument with a default value.

Instructions

Finish the template for the interpret_all() function:

• Make return_sum an optional argument, that is TRUE by default.
• Inside the for loop, iterate over all views: on every iteration, add the result of interpret(v) to count. Remember that interpret(v) returns v for popular days, and 0 otherwise. At the same time, interpret(v) will also do some printouts.
• Finish the if construct:
• If return_sum is TRUE, return count.
• Else, return NULL.

Call this newly defined function on both linkedin and facebook.

# The linkedin and facebook vectors have already been created for you
linkedin <- c(16, 9, 13, 5, 2, 17, 14)
facebook <- c(17, 7, 5, 16, 8, 13, 14)

# The interpret() can be used inside interpret_all()
interpret <- function(num_views) {
  if (num_views > 15) {
    print("You're popular!")
    return(num_views)
  } else {
    print("Try to be more visible!")
    return(0)
  }
}

# Define the interpret_all() function
# views: vector with data to interpret
# return_sum: return total number of views on popular days?
interpret_all <- function(views, return_sum = TRUE) {
  count <- 0

  for (v in views) {
    count <- count + interpret(v)
  }

  if (return_sum == TRUE) {
    return(count)
  } else {
    return(NULL)
  }
}

# Call the interpret_all() function on both linkedin and facebook
interpret_all(linkedin)

## [1] "You're popular!"
## [1] "Try to be more visible!"
## [1] "Try to be more visible!"
## [1] "Try to be more visible!"
## [1] "Try to be more visible!"
## [1] "You're popular!"
## [1] "Try to be more visible!"

## [1] 33

interpret_all(facebook)

## [1] "You're popular!"
## [1] "Try to be more visible!"
## [1] "Try to be more visible!"
## [1] "You're popular!"
## [1] "Try to be more visible!"
## [1] "Try to be more visible!"
## [1] "Try to be more visible!"

## [1] 33

Exercise

There are basically two extremely important functions when it comes down to R packages:

• install.packages(), which as you can expect, installs a given package.
• library() which loads packages, i.e. attaches them to the search list on your R workspace.

To install packages, you need administrator privileges. This means that install.packages() will thus not work in the DataCamp interface. However, almost all CRAN packages are installed on our servers. You can load them with library().

In this exercise, you’ll be learning how to load the ggplot2 package, a powerful package for data visualization. You’ll use it to create a plot of two variables of the mtcars data frame. The data has already been prepared for you in the workspace.

Before starting, execute the following commands in the console:

• search(), to look at the currently attached packages and
• qplot(mtcars$wt, mtcars$hp), to build a plot of two variables of the mtcars data frame.

An error should occur, because you haven’t loaded the ggplot2 package yet!

Instructions

• To fix the error you saw in the console, load the ggplot2 package.
• Now, retry calling the qplot() function with the same arguments.
• Finally, check out the currently attached packages again.

# Load the ggplot2 package
library(ggplot2)

# Retry the qplot() function
qplot(mtcars$wt, mtcars$hp)

# Check out the currently attached packages again
search()

Possible Answers

• Only (1)
• (1) and (2)
• (1), (2) and (3)
• All of them are valid

Exercise

Before you go about solving the exercises below, have a look at the documentation of the lapply() function. The Usage section shows the following expression:

lapply(X, FUN, ...)

To put it generally, lapply takes a vector or list X, and applies the function FUN to each of its members. If FUN requires additional arguments, you pass them after you’ve specified X and FUN (...). The output of lapply() is a list, the same length as X, where each element is the result of applying FUN on the corresponding element of X.

Now that you are truly brushing up on your data science skills, let’s revisit some of the most relevant figures in data science history. We’ve compiled a vector of famous mathematicians/statisticians and the year they were born. Up to you to extract some information!

Instructions

• Have a look at the strsplit() calls, that splits the strings in pioneers on the : sign. The result, split_math is a list of 4 character vectors: the first vector element represents the name, the second element the birth year.
• Use lapply() to convert the character vectors in split_math to lowercase letters: apply tolower() on each of the elements in split_math. Assign the result, which is a list, to a new variable split_low.
• Finally, inspect the contents of split_low with str().

# The vector pioneers has already been created for you
pioneers <- c("GAUSS:1777", "BAYES:1702", "PASCAL:1623", "PEARSON:1857")

# Split names from birth year
split_math <- strsplit(pioneers, split = ":")

# Convert to lowercase strings: split_low
split_low <- lapply(split_math, tolower)

# Take a look at the structure of split_low
str(split_low)

## List of 4
##  $ : chr [1:2] "gauss" "1777"
##  $ : chr [1:2] "bayes" "1702"
##  $ : chr [1:2] "pascal" "1623"
##  $ : chr [1:2] "pearson" "1857"

Exercise

As Filip explained in the instructional video, you can use lapply() on your own functions as well. You just need to code a new function and make sure it is available in the workspace. After that, you can use the function inside lapply() just as you did with base R functions.

In the previous exercise you already used lapply() once to convert the information about your favorite pioneering statisticians to a list of vectors composed of two character strings. Let’s write some code to select the names and the birth years separately.

The sample code already includes code that defined select_first(), that takes a vector as input and returns the first element of this vector.

Instructions

• Apply select_first() over the elements of split_low with lapply() and assign the result to a new variable names.
• Next, write a function select_second() that does the exact same thing for the second element of an inputted vector.
• Finally, apply the select_second() function over split_low and assign the output to the variable years.

# Code from previous exercise:
pioneers <- c("GAUSS:1777", "BAYES:1702", "PASCAL:1623", "PEARSON:1857")
split <- strsplit(pioneers, split = ":")
split_low <- lapply(split, tolower)

# Write function select_first()
select_first <- function(x) {
  x[1]
}

# Apply select_first() over split_low: names
names <- lapply(split_low, select_first)

# Write function select_second()
select_second <- function(x) {
  x[2]
}

# Apply select_second() over split_low: years
years <- lapply(split_low, select_second)

Exercise

Writing your own functions and then using them inside lapply() is quite an accomplishment! But defining functions to use them only once is kind of overkill, isn’t it? That’s why you can use so-called anonymous functions in R.

Previously, you learned that functions in R are objects in their own right. This means that they aren’t automatically bound to a name. When you create a function, you can use the assignment operator to give the function a name. It’s perfectly possible, however, to not give the function a name. This is called an anonymous function:

# Named function
triple <- function(x) { 3 * x }

# Anonymous function with same implementation
function(x) { 3 * x }

# Use anonymous function inside lapply()
lapply(list(1,2,3), function(x) { 3 * x })

Instructions

• Transform the first call of lapply() such that it uses an anonymous function that does the same thing.
• In a similar fashion, convert the second call of lapply to use an anonymous version of the select_second() function.
• Remove both the definitions of select_first() and select_second(), as they are no longer useful.

# Definition of split_low
pioneers <- c("GAUSS:1777", "BAYES:1702", "PASCAL:1623", "PEARSON:1857")
split <- strsplit(pioneers, split = ":")
split_low <- lapply(split, tolower)

# Transform: use anonymous function inside lapply
names <- lapply(split_low, function(x) { x[1] })

# Transform: use anonymous function inside lapply
years <- lapply(split_low, function(x) { x[2] })

Exercise

In the video, the triple() function was transformed to the multiply() function to allow for a more generic approach. lapply() provides a way to handle functions that require more than one argument, such as the multiply() function:

multiply <- function(x, factor) {
  x * factor
}
lapply(list(1,2,3), multiply, factor = 3)

On the right we’ve included a generic version of the select functions that you’ve coded earlier: select_el(). It takes a vector as its first argument, and an index as its second argument. It returns the vector’s element at the specified index.

Instructions

Use lapply() twice to call select_el() over all elements in split_low: once with the index equal to 1 and a second time with the index equal to 2. Assign the result to names and years, respectively.

# Definition of split_low
pioneers <- c("GAUSS:1777", "BAYES:1702", "PASCAL:1623", "PEARSON:1857")
split <- strsplit(pioneers, split = ":")
split_low <- lapply(split, tolower)

# Generic select function
select_el <- function(x, index) {
  x[index]
}

# Use lapply() twice on split_low: names and years
names <- lapply(split_low, select_el, index = 1)
years <- lapply(split_low, select_el, index = 2)

Possible Answers

• list(NULL, NULL, "1623", "1857")
• list("gauss", "bayes", NULL, NULL)
• list(“1777”, “1702”, NULL, NULL)
• list("1777", "1702")

Wonderful! Feel free to experiment some more with your code in the console. Did you notice that lapply() always returns a list, no matter the input? This can be kind of annoying. In the next video tutorial you’ll learn about sapply() to solve this.

Exercise

You can use sapply() similar to how you used lapply(). The first argument of sapply() is the list or vector X over which you want to apply a function, FUN. Potential additional arguments to this function are specified afterwards (...):

sapply(X, FUN, ...)

In the next couple of exercises, you’ll be working with the variable temp, that contains temperature measurements for 7 days. temp is a list of length 7, where each element is a vector of length 5, representing 5 measurements on a given day. This variable has already been defined in the workspace: type str(temp) to see its structure.

Instructions

• Use lapply() to calculate the minimum (built-in function min()) of the temperature measurements for every day.
• Do the same thing but this time with sapply(). See how the output differs.
• Use lapply() to compute the the maximum (max()) temperature for each day.
• Again, use sapply() to solve the same question and see how lapply() and sapply() differ.

# temp has already been defined in the workspace
temp <- list(c(3, 7, 9, 6, -1), c(6, 9, 12, 13, 5), c(4, 8, 3, -1, -3), c(1, 4, 7, 2, -2), c(5, 7, 9, 4, 2), c(-3, 5, 8, 9, 4), c(3, 6, 9, 4, 1))

# Use lapply() to find each day's minimum temperature
lapply(temp, min)

## [[1]]
## [1] -1
## 
## [[2]]
## [1] 5
## 
## [[3]]
## [1] -3
## 
## [[4]]
## [1] -2
## 
## [[5]]
## [1] 2
## 
## [[6]]
## [1] -3
## 
## [[7]]
## [1] 1

# Use sapply() to find each day's minimum temperature
sapply(temp, min)

## [1] -1  5 -3 -2  2 -3  1

# Use lapply() to find each day's maximum temperature
lapply(temp, max)

## [[1]]
## [1] 9
## 
## [[2]]
## [1] 13
## 
## [[3]]
## [1] 8
## 
## [[4]]
## [1] 7
## 
## [[5]]
## [1] 9
## 
## [[6]]
## [1] 9
## 
## [[7]]
## [1] 9

# Use sapply() to find each day's maximum temperature
sapply(temp, max)

## [1]  9 13  8  7  9  9  9

Nice! Can you tell the difference between the output of lapply() and sapply()? The former returns a list, while the latter returns a vector that is a simplified version of this list. Notice that this time, unlike in the cities example of the instructional video, the vector is not named.

Exercise

Like lapply(), sapply() allows you to use self-defined functions and apply them over a vector or a list:

sapply(X, FUN, ...)

Here, FUN can be one of R’s built-in functions, but it can also be a function you wrote. This self-written function can be defined before hand, or can be inserted directly as an anonymous function.

Instructions

• Finish the definition of extremes_avg(): it takes a vector of temperatures and calculates the average of the minimum and maximum temperatures of the vector.
• Next, use this function inside sapply() to apply it over the vectors inside temp. Use the same function over temp with lapply() and see how the outputs differ.

# temp is already defined in the workspace
temp

## [[1]]
## [1]  3  7  9  6 -1
## 
## [[2]]
## [1]  6  9 12 13  5
## 
## [[3]]
## [1]  4  8  3 -1 -3
## 
## [[4]]
## [1]  1  4  7  2 -2
## 
## [[5]]
## [1] 5 7 9 4 2
## 
## [[6]]
## [1] -3  5  8  9  4
## 
## [[7]]
## [1] 3 6 9 4 1

# Finish function definition of extremes_avg
extremes_avg <- function(x) {
  ( min(x) + max(x) ) / 2
}

# Apply extremes_avg() over temp using sapply()
sapply(temp, extremes_avg)

## [1] 4.0 9.0 2.5 2.5 5.5 3.0 5.0

# Apply extremes_avg() over temp using lapply()
lapply(temp, extremes_avg)

## [[1]]
## [1] 4
## 
## [[2]]
## [1] 9
## 
## [[3]]
## [1] 2.5
## 
## [[4]]
## [1] 2.5
## 
## [[5]]
## [1] 5.5
## 
## [[6]]
## [1] 3
## 
## [[7]]
## [1] 5

Great job! Of course, you could have solved this exercise using an anonymous function, but this would require you to use the code inside the definition of extremes_avg() twice. Duplicating code should be avoided as much as possible!

Exercise

In the previous exercises, you’ve seen how sapply() simplifies the list that lapply() would return by turning it into a vector. But what if the function you’re applying over a list or a vector returns a vector of length greater than 1? If you don’t remember from the video, don’t waste more time in the valley of ignorance and head over to the instructions!

Instructions

Finish the definition of the extremes() function. It takes a vector of numerical values and returns a vector containing the minimum and maximum values of a given vector, with the names “min” and “max”, respectively. * Apply this function over the vector temp using sapply(). * Finally, apply this function over the vector temp using lapply() as well.

# temp is already available in the workspace

# Create a function that returns min and max of a vector: extremes
extremes <- function(x) {
  c(min = min(x), max = max(x))
}

# Apply extremes() over temp with sapply()
sapply(temp, extremes)

##     [,1] [,2] [,3] [,4] [,5] [,6] [,7]
## min   -1    5   -3   -2    2   -3    1
## max    9   13    8    7    9    9    9

# Apply extremes() over temp with lapply()
lapply(temp, extremes)

## [[1]]
## min max 
##  -1   9 
## 
## [[2]]
## min max 
##   5  13 
## 
## [[3]]
## min max 
##  -3   8 
## 
## [[4]]
## min max 
##  -2   7 
## 
## [[5]]
## min max 
##   2   9 
## 
## [[6]]
## min max 
##  -3   9 
## 
## [[7]]
## min max 
##   1   9

Wonderful! Have a final look at the console and see how sapply() did a great job at simplifying the rather uninformative ‘list of vectors’ that lapply() returns. It actually returned a nicely formatted matrix!

Exercise

It seems like we’ve hit the jackpot with sapply(). On all of the examples so far, sapply() was able to nicely simplify the rather bulky output of lapply(). But, as with life, there are things you can’t simplify. How does sapply() react?

We already created a function, below_zero(), that takes a vector of numerical values and returns a vector that only contains the values that are strictly below zero.

Instructions

• Apply below_zero() over temp using sapply() and store the result in freezing_s.
• Apply below_zero() over temp using lapply(). Save the resulting list in a variable freezing_l.
• Compare freezing_s to freezing_l using the identical() function.

# temp is already prepared for you in the workspace

# Definition of below_zero()
below_zero <- function(x) {
  return(x[x < 0])
}

# Apply below_zero over temp using sapply(): freezing_s
freezing_s <- sapply(temp, below_zero)

# Apply below_zero over temp using lapply(): freezing_l
freezing_l <- lapply(temp, below_zero)

# Are freezing_s and freezing_l identical?
identical(freezing_s, freezing_l)

## [1] TRUE

Nice one! Given that the length of the output of below_zero() changes for different input vectors, sapply() is not able to nicely convert the output of lapply() to a nicely formatted matrix. Instead, the output values of sapply() and lapply() are exactly the same, as shown by the TRUE output of identical().

Exercise

You already have some apply tricks under your sleeve, but you’re surely hungry for some more, aren’t you? In this exercise, you’ll see how sapply() reacts when it is used to apply a function that returns NULL over a vector or a list.

A function print_info(), that takes a vector and prints the average of this vector, has already been created for you. It uses the cat() function.

Instructions

• Apply print_info() over the contents of temp with sapply().
• Repeat this process with lapply(). Do you notice the difference?

# temp is already available in the workspace

# Definition of print_info()
print_info <- function(x) {
  cat("The average temperature is", mean(x), "\n")
}

# Apply print_info() over temp using sapply()
sapply(temp, print_info)

## The average temperature is 4.8 
## The average temperature is 9 
## The average temperature is 2.2 
## The average temperature is 2.4 
## The average temperature is 5.4 
## The average temperature is 4.6 
## The average temperature is 4.6

## [[1]]
## NULL
## 
## [[2]]
## NULL
## 
## [[3]]
## NULL
## 
## [[4]]
## NULL
## 
## [[5]]
## NULL
## 
## [[6]]
## NULL
## 
## [[7]]
## NULL

# Apply print_info() over temp using lapply()
lapply(temp, print_info)

## The average temperature is 4.8 
## The average temperature is 9 
## The average temperature is 2.2 
## The average temperature is 2.4 
## The average temperature is 5.4 
## The average temperature is 4.6 
## The average temperature is 4.6

## [[1]]
## NULL
## 
## [[2]]
## NULL
## 
## [[3]]
## NULL
## 
## [[4]]
## NULL
## 
## [[5]]
## NULL
## 
## [[6]]
## NULL
## 
## [[7]]
## NULL

Great! Notice here that, quite surprisingly, sapply() does not simplify the list of NULL‘s. That’s because the ’vector-version’ of a list of NULL’s would simply be a NULL, which is no longer a vector with the same length as the input. Proceed to the next exercise.

Possible Answers

• 1. and (3)
• (2) and (3)
• 1. and (4)
• (2), (3) and (4)

Exercise

Before you get your hands dirty with the third and last apply function that you’ll learn about in this intermediate R course, let’s take a look at its syntax. The function is called vapply(), and it has the following syntax:

vapply(X, FUN, FUN.VALUE, ..., USE.NAMES = TRUE)

Over the elements inside X, the function FUN is applied. The FUN.VALUE argument expects a template for the return argument of this function FUN. USE.NAMES is TRUE by default; in this case vapply() tries to generate a named array, if possible.

For the next set of exercises, you’ll be working on the temp list again, that contains 7 numerical vectors of length 5. We also coded a function basics() that takes a vector, and returns a named vector of length 3, containing the minimum, mean and maximum value of the vector respectively.

Instructions

Apply the function basics() over the list of temperatures, temp, using vapply(). This time, you can use numeric(3) to specify the FUN.VALUE argument.

# temp is already available in the workspace

# Definition of basics()
basics <- function(x) {
  c(min = min(x), mean = mean(x), max = max(x))
}

# Apply basics() over temp using vapply()
vapply(temp, basics, numeric(3))

##      [,1] [,2] [,3] [,4] [,5] [,6] [,7]
## min  -1.0    5 -3.0 -2.0  2.0 -3.0  1.0
## mean  4.8    9  2.2  2.4  5.4  4.6  4.6
## max   9.0   13  8.0  7.0  9.0  9.0  9.0

Perfect! Notice how, just as with sapply(), vapply() neatly transfers the names that you specify in the basics() function to the row names of the matrix that it returns.

Exercise

So far you’ve seen that vapply() mimics the behavior of sapply() if everything goes according to plan. But what if it doesn’t?

In the video, Filip showed you that there are cases where the structure of the output of the function you want to apply, FUN, does not correspond to the template you specify in FUN.VALUE. In that case, vapply() will throw an error that informs you about the misalignment between expected and actual output.

Instructions

• Inspect the code on the right and try to run it. If you haven’t changed anything, an error should pop up. That’s because vapply() still expects basics() to return a vector of length 3. The error message gives you an indication of what’s wrong.
• Try to fix the error by editing the vapply() command.

# temp is already available in the workspace

# Definition of the basics() function
basics <- function(x) {
  c(min = min(x), mean = mean(x), median = median(x), max = max(x))
}

# Fix the error:
vapply(temp, basics, numeric(4))

##        [,1] [,2] [,3] [,4] [,5] [,6] [,7]
## min    -1.0    5 -3.0 -2.0  2.0 -3.0  1.0
## mean    4.8    9  2.2  2.4  5.4  4.6  4.6
## median  6.0    9  3.0  2.0  5.0  5.0  4.0
## max     9.0   13  8.0  7.0  9.0  9.0  9.0

Exercise

As highlighted before, vapply() can be considered a more robust version of sapply(), because you explicitly restrict the output of the function you want to apply. Converting your sapply() expressions in your own R scripts to vapply() expressions is therefore a good practice (and also a breeze!).

Instructions

Convert all the sapply() expressions on the right to their vapply() counterparts. Their results should be exactly the same; you’re only adding robustness. You’ll need the templates numeric(1) and logical(1).

# temp is already defined in the workspace

# Convert to vapply() expression
vapply(temp, max, numeric(1))

## [1]  9 13  8  7  9  9  9

# Convert to vapply() expression
vapply(temp, function(x, y) { mean(x) > y }, y = 5, logical(1))

## [1] FALSE  TRUE FALSE FALSE  TRUE FALSE FALSE

Exercise

Have another look at some useful math functions that R features:

• abs(): Calculate the absolute value.
• sum(): Calculate the sum of all the values in a data structure.
• mean(): Calculate the arithmetic mean.
• round(): Round the values to 0 decimal places by default. Try out ?round in the console for variations of round() and ways to change the number of digits to round to.

As a data scientst in training, you’ve estimated a regression model on the sales data for the past six months. After evaluating your model, you see that the training error of your model is quite regular, showing both positive and negative values. The error values are already defined in the workspace on the right (errors).

Instructions

Calculate the sum of the absolute rounded values of the training errors. You can work in parts, or with a single one-liner. There’s no need to store the result in a variable, just have R print it.

# The errors vector has already been defined for you
errors <- c(1.9, -2.6, 4.0, -9.5, -3.4, 7.3)

# Sum of absolute rounded values of errors
round(errors, digits = 0)

## [1]   2  -3   4 -10  -3   7

abs(errors)

## [1] 1.9 2.6 4.0 9.5 3.4 7.3

sum(abs(round(errors, digits = 0)))

## [1] 29

Exercise

We went ahead and included some code on the right, but there’s still an error. Can you trace it and fix it?

In times of despair, help with functions such as sum() and rev() are a single command away; simply use ?sum and ?rev in the console.

Instructions

Fix the error by including code on the last line. Remember: you want to call mean() only once!

# Don't edit these two lines
vec1 <- c(1.5, 2.5, 8.4, 3.7, 6.3)
vec2 <- rev(vec1)

# Fix the error
mean(abs(vec1), abs(vec2),trim = 0)

## Warning in if (na.rm) x <- x[!is.na(x)]: the condition has length > 1 and
## only the first element will be used

## [1] 4.48

Nice work! If you check out the documentation of mean(), you’ll see that only the first argument, x, should be a vector. If you also specify a second argument, R will match the arguments by position and expect a specification of the trim argument. Therefore, merging the two vectors is a must!

Exercise

R features a bunch of functions to juggle around with data structures::

• seq(): Generate sequences, by specifying the from, to, and by arguments.
• rep(): Replicate elements of vectors and lists.
• sort(): Sort a vector in ascending order. Works on numerics, but also on character strings and logicals.
• rev(): Reverse the elements in a data structures for which reversal is defined.
• str(): Display the structure of any R object.
• append(): Merge vectors or lists.
• is.*(): Check for the class of an R object.
• as.*(): Convert an R object from one class to another.
• unlist(): Flatten (possibly embedded) lists to produce a vector. Remember the social media profile views data? Your LinkedIn and Facebook view counts for the last seven days are already defined as lists on the right.

Instructions

• Convert both linkedin and facebook lists to a vector, and store them as li_vec and fb_vec respectively.
• Next, append fb_vec to the li_vec (Facebook data comes last). Save the result as social_vec.
• Finally, sort social_vec from high to low. Print the resulting vector.

# The linkedin and facebook lists have already been created for you
linkedin <- list(16, 9, 13, 5, 2, 17, 14)
facebook <- list(17, 7, 5, 16, 8, 13, 14)

# Convert linkedin and facebook to a vector: li_vec and fb_vec
li_vec <- unlist(linkedin)
fb_vec <- unlist(facebook)

# Append fb_vec to li_vec: social_vec
social_vec <- append(li_vec, fb_vec)

# Sort social_vec
sort(social_vec, decreasing = TRUE)

##  [1] 17 17 16 16 14 14 13 13  9  8  7  5  5  2

Exercise

Just as before, let’s switch roles. It’s up to you to see what unforgivable mistakes we’ve made. Go fix them!

Instructions

Correct the expression. Make sure that your fix still uses the functions rep() and seq().

# Fix me
rep(seq(1, 7, by = 2), times = 7)

##  [1] 1 3 5 7 1 3 5 7 1 3 5 7 1 3 5 7 1 3 5 7 1 3 5 7 1 3 5 7

Exercise

There is a popular story about young Gauss. As a pupil, he had a lazy teacher who wanted to keep the classroom busy by having them add up the numbers 1 to 100. Gauss came up with an answer almost instantaneously, 5050. On the spot, he had developed a formula for calculating the sum of an arithmetic series. There are more general formulas for calculating the sum of an arithmetic series with different starting values and increments. Instead of deriving such a formula, why not use R to calculate the sum of a sequence?

Instructions

• Using the function seq(), create a sequence that ranges from 1 to 500 in increments of 3. Assign the resulting vector to a variable seq1.
• Again with the function seq(), create a sequence that ranges from 1200 to 900 in increments of -7. Assign it to a variable seq2.
• Calculate the total sum of the sequences, either by using the sum() function twice and adding the two results, or by first concatenating the sequences and then using the sum() function once. Print the result to the console.

# Create first sequence: seq1
seq1 <- seq(from = 1, to = 500, by = 3)

# Create second sequence: seq2
seq2 <- seq(from = 1200, to = 900, by = -7)

# Calculate total sum of the sequences
sum(seq1, seq2)

## [1] 87029

Exercise

In their most basic form, regular expressions can be used to see whether a pattern exists inside a character string or a vector of character strings. For this purpose, you can use:

• grepl(), which returns TRUE when a pattern is found in the corresponding character string.
• grep(), which returns a vector of indices of the character strings that contains the pattern.

Both functions need a pattern and an x argument, where pattern is the regular expression you want to match for, and the x argument is the character vector from which matches should be sought.

In this and the following exercises, you’ll be querying and manipulating a character vector of email addresses! The vector emails has already been defined in the editor on the right so you can begin with the instructions straight away!

Instructions

• Use grepl() to generate a vector of logicals that indicates whether these email addressess contain "edu". Print the result to the output.
• Do the same thing with grep(), but this time save the resulting indexes in a variable hits.
• Use the variable hits to select from the emails vector only the emails that contain "edu".

# The emails vector has already been defined for you
emails <- c("john.doe@ivyleague.edu", "education@world.gov", "dalai.lama@peace.org",
            "invalid.edu", "quant@bigdatacollege.edu", "cookie.monster@sesame.tv")

# Use grepl() to match for "edu"
grepl(pattern = "edu", x = emails)

## [1]  TRUE  TRUE FALSE  TRUE  TRUE FALSE

# Use grep() to match for "edu", save result to hits
hits <- grep(pattern = "edu", x = emails)

# Subset emails using hits
emails[hits]

## [1] "john.doe@ivyleague.edu"   "education@world.gov"     
## [3] "invalid.edu"              "quant@bigdatacollege.edu"

Bellissimo! You can probably guess what we’re trying to achieve here: select all the emails that end with “.edu”. However, the strings education@world.gov and invalid.edu were also matched. Let’s see in the next exercise what you can do to improve our pattern and remove these false positives.

Exercise

You can use the caret, ^, and the dollar sign, $ to match the content located in the start and end of a string, respectively. This could take us one step closer to a correct pattern for matching only the “.edu” email addresses from our list of emails. But there’s more that can be added to make the pattern more robust:

• @, because a valid email must contain an at-sign.
• .*, which matches any character (.) zero or more times (*). Both the dot and the asterisk are metacharacters. You can use them to match any character between the at-sign and the “.edu” portion of an email address.
• \\.edu$, to match the “.edu” part of the email at the end of the string. The \\ part escapes the dot: it tells R that you want to use the . as an actual character.

Instructions

• Use grepl() with the more advanced regular expression to return a logical vector. Simply print the result.
• Do a similar thing with grep() to create a vector of indices. Store the result in the variable hits.
• Use emails[hits] again to subset the emails vector.

# The emails vector has already been defined for you
emails <- c("john.doe@ivyleague.edu", "education@world.gov", "dalai.lama@peace.org",
            "invalid.edu", "quant@bigdatacollege.edu", "cookie.monster@sesame.tv")

# Use grepl() to match for .edu addresses more robustly
grepl(pattern = "@.*\\.edu", x = emails)

## [1]  TRUE FALSE FALSE FALSE  TRUE FALSE

# Use grep() to match for .edu addresses more robustly, save result to hits
hits <- grep(pattern = "@.*\\.edu", x = emails)

# Subset emails using hits
emails[hits]

## [1] "john.doe@ivyleague.edu"   "quant@bigdatacollege.edu"

Great! A careful construction of our regular expression leads to more meaningful matches. However, even our robust email selector will often match some incorrect email addresses (for instance kiara@@fakemail.edu). Let’s not worry about this too much and continue with sub() and gsub() to actually edit the email addresses!

Exercise

While grep() and grepl() were used to simply check whether a regular expression could be matched with a character vector, sub() and gsub() take it one step further: you can specify a replacement argument. If inside the character vector x, the regular expression pattern is found, the matching element(s) will be replaced with replacement. sub() only replaces the first match, whereas gsub() replaces all matches.

Suppose that emails vector you’ve been working with is an excerpt of DataCamp’s email database. Why not offer the owners of the .edu email addresses a new email address on the datacamp.edu domain? This could be quite a powerful marketing stunt: Online education is taking over traditional learning institutions! Convert your email and be a part of the new generation!

Instructions

With the advanced regular expression "@.*\\.edu$", use sub() to replace the match with "@datacamp.edu". Since there will only be one match per character string, gsub() is not necessary here. Inspect the resulting output.

# The emails vector has already been defined for you
emails <- c("john.doe@ivyleague.edu", "education@world.gov", "global@peace.org",
            "invalid.edu", "quant@bigdatacollege.edu", "cookie.monster@sesame.tv")

# Use sub() to convert the email domains to datacamp.edu
sub(pattern = "@.*\\.edu$", replacement = "@datacamp.edu", x = emails)

## [1] "john.doe@datacamp.edu"    "education@world.gov"     
## [3] "global@peace.org"         "invalid.edu"             
## [5] "quant@datacamp.edu"       "cookie.monster@sesame.tv"

Awesome! Notice how only the valid .edu addresses are changed while the other emails remain unchanged. To get a taste of other things you can accomplish with regex, head over to the next exercise.

Possible Answers

• A vector of integers containing: 1, 24, 2, 3, 2, 1.
• The vector awards gets returned as there isn’t a single element in awards that matches the regular expression.
• A vetor of character strings containing “1”, “24”, “2”, “3”, “2”, “1”.
• A vector of character strings containing “Won 1 Oscar.”, “24”, “2”, “3”, “2”, “1”.

Great! Can you explain why all of this happened? The ([0-9]+) selects the entire number that comes before the word “nomination” in the string, and the entire match gets replaced by this number because of the \\1 that reference to the content inside the parentheses. The next video will get you up to speed with times and dates in R!

Exercise

In R, dates are represented by Date objects, while times are represented by POSIXct objects. Under the hood, however, these dates and times are simple numerical values. Date objects store the number of days since the 1st of January in 1970. POSIXct objects on the other hand, store the number of seconds since the 1st of January in 1970.

The 1st of January in 1970 is the common origin for representing times and dates in a wide range of programming languages. There is no particular reason for this; it is a simple convention. Of course, it’s also possible to create dates and times before 1970; the corresponding numerical values are simply negative in this case.

Instructions

• Ask R for the current date, and store the result in a variable today.
• To see what today looks like under the hood, call unclass() on it.
• Ask R for the current time, and store the result in a variable, now.
• To see the numerical value that corresponds to now, call unclass() on it.

# Get the current date: today
today <- Sys.Date()
today

## [1] "2018-09-03"

# See what today looks like under the hood
unclass(today)

## [1] 17777

# Get the current time: now
now <- Sys.time()
now

## [1] "2018-09-03 20:24:52 +08"

# See what now looks like under the hood
unclass(now)

## [1] 1535977492

Exercise

To create a Date object from a simple character string in R, you can use the as.Date() function. The character string has to obey a format that can be defined using a set of symbols (the examples correspond to 13 January, 1982):

• %Y: 4-digit year (1982)
• %y: 2-digit year (82)
• %m: 2-digit month (01)
• %d: 2-digit day of the month (13)
• %A: weekday (Wednesday)
• %a: abbreviated weekday (Wed)
• %B: month (January)
• %b: abbreviated month (Jan)

The following R commands will all create the same Date object for the 13th day in January of 1982:

as.Date("1982-01-13")
as.Date("Jan-13-82", format = "%b-%d-%y")
as.Date("13 January, 1982", format = "%d %B, %Y")

Notice that the first line here did not need a format argument, because by default R matches your character string to the formats "%Y-%m-%d" or "%Y/%m/%d".

In addition to creating dates, you can also convert dates to character strings that use a different date notation. For this, you use the format() function. Try the following lines of code:

today <- Sys.Date()
format(Sys.Date(), format = "%d %B, %Y")
format(Sys.Date(), format = "Today is a %A!")

Instructions

• In the editor on the right, three character strings representing dates have been created. Convert them to dates using as.Date(), and assign them to date1, date2, and date3 respectively. The code for date1 is already included.
• Extract useful information from the dates as character strings using format(). From the first date, select the weekday. From the second date, select the day of the month. From the third date, you should select the abbreviated month and the 4-digit year, separated by a space.

# Definition of character strings representing dates
str1 <- "May 23, '96"
str2 <- "2012-03-15"
str3 <- "30/January/2006"

# Convert the strings to dates: date1, date2, date3
date1 <- as.Date(str1, format = "%b %d, '%y")
date2 <- as.Date(str2, format = "%Y-%m-%d")
date3 <- as.Date(str3, format = "%d/%B/%Y")

# Convert dates to formatted strings
format(date1, "%A")

## [1] "Thursday"

format(date2, "%d")

## [1] "15"

format(date3, "%b %Y")

## [1] "Jan 2006"

Exercise

Similar to working with dates, you can use as.POSIXct() to convert from a character string to a POSIXct object, and format() to convert from a POSIXct object to a character string. Again, you have a wide variety of symbols:

• %H: hours as a decimal number (00-23)
• %I: hours as a decimal number (01-12)
• %M: minutes as a decimal number
• %S: seconds as a decimal number
• %T: shorthand notation for the typical format %H:%M:%S
• %p: AM/PM indicator

For a full list of conversion symbols, consult the strptime documentation in the console:

?strptime

Again, as.POSIXct() uses a default format to match character strings. In this case, it’s %Y-%m-%d %H:%M:%S. In this exercise, abstraction is made of different time zones.

Instructions

• Convert two strings that represent timestamps, str1 and str2, to POSIXct objects called time1 and time2.
• Using format(), create a string from time1 containing only the minutes.
• From time2, extract the hours and minutes as “hours:minutes AM/PM”. Refer to the assignment text above to find the correct conversion symbols!

# Definition of character strings representing times
str1 <- "May 23, '96 hours:23 minutes:01 seconds:45"
str2 <- "2012-3-12 14:23:08"

# Convert the strings to POSIXct objects: time1, time2
time1 <- as.POSIXct(str1, format = "%B %d, '%y hours:%H minutes:%M seconds:%S")
time2 <- as.POSIXct(str2, format = "%Y-%m-%d %H:%M:%S")

# Convert times to formatted strings
format(time1, "%M")

## [1] "01"

format(time2, "%I:%M %p")

## [1] "02:23 PM"

Exercise

Both Date and POSIXct R objects are represented by simple numerical values under the hood. This makes calculation with time and date objects very straightforward: R performs the calculations using the underlying numerical values, and then converts the result back to human-readable time information again.

You can increment and decrement Date objects, or do actual calculations with them (try it out in the console!):

today <- Sys.Date()
today + 1
today - 1

as.Date("2015-03-12") - as.Date("2015-02-27")

To control your eating habits, you decided to write down the dates of the last five days that you ate pizza. In the workspace, these dates are defined as five Date objects, day1 to day5. The code on the right also contains a vector pizza with these 5 Date objects.

Instructions

• Calculate the number of days that passed between the last and the first day you ate pizza. Print the result.
• Use the function diff() on pizza to calculate the differences between consecutive pizza days. Store the result in a new variable day_diff.
• Calculate the average period between two consecutive pizza days. Print the result.

# day1, day2, day3, day4 and day5 are already available in the workspace
day1 <- as.Date("2018-08-15")
day2 <- as.Date("2018-08-17")
day3 <- as.Date("2018-08-22")
day4 <- as.Date("2018-08-28")
day5 <- as.Date("2018-09-02")

# Difference between last and first pizza day
as.Date(day5) - as.Date(day1)

## Time difference of 18 days

# Create vector pizza
pizza <- c(day1, day2, day3, day4, day5)

# Create differences between consecutive pizza days: day_diff
day_diff <- diff(pizza)

# Average period between two consecutive pizza days
mean(day_diff)

## Time difference of 4.5 days

Exercise

Calculations using POSIXct objects are completely analogous to those using Date objects. Try to experiment with this code to increase or decrease POSIXct objects:

now <- Sys.time()
now + 3600          # add an hour
now - 3600 * 24     # subtract a day

Adding or substracting time objects is also straightforward:

birth <- as.POSIXct("1879-03-14 14:37:23")
death <- as.POSIXct("1955-04-18 03:47:12")
einstein <- death - birth
einstein

You’re developing a website that requires users to log in and out. You want to know what is the total and average amount of time a particular user spends on your website. This user has logged in 5 times and logged out 5 times as well. These times are gathered in the vectors login and logout, which are already defined in the workspace.

Instructions

• Calculate the difference between the two vectors logout and login, i.e. the time the user was online in each independent session. Store the result in a variable time_online.
• Inspect the variable time_online by printing it.
• Calculate the total time that the user was online. Print the result.
• Calculate the average time the user was online. Print the result.

# login and logout are already defined in the workspace
login <- as.POSIXct(c("2018-08-19 10:18:04 UTC", "2018-08-24 09:14:18 UTC"
, "2018-08-24 12:21:51 UTC", "2018-08-24 12:37:24 UTC"
, "2018-08-26 21:37:55 UTC"))
logout <- as.POSIXct(c("2018-08-19 10:56:29 UTC", "2018-08-24 09:14:52 UTC"
, "2018-08-24 12:35:48 UTC", "2018-08-24 13:17:22 UTC"
, "2018-08-26 22:08:47 UTC"))

# Calculate the difference between login and logout: time_online
time_online = logout - login

# Inspect the variable time_online
time_online

## Time differences in secs
## [1] 2305   34  837 2398 1852

# Calculate the total time online
sum(time_online)

## Time difference of 7426 secs

# Calculate the average time online
mean(time_online)

## Time difference of 1485.2 secs

Exercise

The dates when a season begins and ends can vary depending on who you ask. People in Australia will tell you that spring starts on September 1st. The Irish people in the Northern hemisphere will swear that spring starts on February 1st, with the celebration of St. Brigid’s Day. Then there’s also the difference between astronomical and meteorological seasons: while astronomers are used to equinoxes and solstices, meteorologists divide the year into 4 fixed seasons that are each three months long. (source: www.timeanddate.com)

A vector astro, which contains character strings representing the dates on which the 4 astronomical seasons start, has been defined on your workspace. Similarly, a vector meteo has already been created for you, with the meteorological beginnings of a season.

Instructions

• Use as.Date() to convert the astro vector to a vector containing Date objects. You will need the %d, %b and %Y symbols to specify the format. Store the resulting vector as astro_dates.
• Use as.Date() to convert the meteo vector to a vector with Date objects. This time, you will need the %B, %d and %y symbols for the format argument. Store the resulting vector as meteo_dates.
• With a combination of max(), abs() and -, calculate the maximum absolute difference between the astronomical and the meteorological beginnings of a season, i.e. astro_dates and meteo_dates. Simply print this maximum difference to the console output.

astro <- c("20-Mar-2015", "25-Jun-2015", "23-Sep-2015", "22-Dec-2015")
names(astro) <- c("spring", "summer", "fall", "winter")

meteo <- c("March 1, 15", "June 1, 15", "September 1, 15", "December 1, 15")
names(meteo) <- c("spring", "summer", "fall", "winter")

# Convert astro to vector of Date objects: astro_dates
astro_dates <- as.Date(astro, format = "%d-%b-%Y")

# Convert meteo to vector of Date objects: meteo_dates
meteo_dates <- as.Date(meteo, format = "%B %d, %y")

# Calculate the maximum absolute difference between astro_dates and meteo_dates
max(abs(astro_dates - meteo_dates))

## Time difference of 24 days