Intermediate R

Course Description

Intermediate R is the next stop on your journey in mastering the R programming language. In this R training, you will learn about conditional statements, loops, and functions to power your own R scripts. Next, make your R code more efficient and readable using the apply functions. Finally, the utilities chapter gets you up to speed with regular expressions in R, data structure manipulations, and times and dates. This course will allow you to take the next step in advancing your overall knowledge and capabilities while programming in R.

Conditionals and Control Flow

In this chapter, you’ll learn about relational operators for comparing R objects, and logical operators like “and” and “or” for combining TRUE and FALSE values. Then, you’ll use this knowledge to build conditional statements.

Equality

The most basic form of comparison is equality. Let’s briefly recap its syntax. The following statements all evaluate to TRUE.

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”.

• 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

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.

Greater and less than

Apart from equality operators, we have 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 is treated as 1 for arithmetic, and FALSE is treated as 0. Therefore, FALSE < TRUE is TRUE.

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

Compare vectors

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.

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

Compare matrices

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.

# 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

& and |

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.

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 >= 20 & last < 15

## [1] FALSE

& and | (2)

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?

• 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 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 <= 14 & views > 11

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

You’ll have noticed how easy it is to use logical operators to vectors and matrices. What do these results tell us? The third day of the recordings was the only day where your LinkedIn profile was visited more than 10 times, while your Facebook profile wasn’t.

Reverse the result: !

On top of the & and | operators, you also learned about the ! operator, which negates a logical value. To refresh your memory, here are some R expressions that use !. They all evaluate to FALSE:

!TRUE

!(5 > 3)

!!FALSE

What would the following set of R expressions return?

x <- 5
y <- 7
!(!(x < 4) & !!!(y > 12))

## [1] FALSE

Blend it all together

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.

• 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

# 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)

The if statement

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.

• Examine the if statement that prints out “Showing LinkedIn information” if the medium variable equals “LinkedIn”.
• Code an if statement that prints “You are 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 are popular!")
}

num_views <- 20

if (num_views > 15) {
  print("You are popular!")
}

## [1] "You are popular!"

Add an else

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!

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 <- "Facebook"
num_views <- 14

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

## [1] "Unknown medium"

# 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!"

Customize further: else if

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!

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") {
  print("Showing Facebook information")
  # Add code to print correct string when condition is TRUE

} 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) {
  print("Your number of views is average")
  # Add code to print correct string when condition is TRUE
} else {
  print("Try to be more visible!")
}

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

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.

Else if 2.0

You can do anything you want inside if-else constructs. You can even put in another set of conditional statements. Examine the following code chunk:

number <- 6

if (number < 10) {
  if (number < 5) {
    result <- "extra small"
  } else {
    result <- "small"
  }
} else if (number < 100) {
  result <- "medium"
} else {
  result <- "large"
}
print(result)

## [1] "small"

Take control!

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.

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.

li <- 15
fb <- 9

if (li > 15 & fb > 15) {
  sms <- 2 * (li + fb)
} else if (li < 10 & fb < 10) {
  sms <- (li + fb)/2
} else {
  sms <- li + fb
}

print(sms)

## [1] 24

Loops

Write a while loop

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.

If your session expires when you run your code, check the body of your while loop carefully.

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 and assign this new value to speed again. This step is crucial; otherwise your while loop will never stop and your session will expire.

# 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

Throw in more conditionals

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?

• 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.

If the session keeps timing out and throwing an error, you are probably stuck in an infinite loop! Check the body of your while loop and make sure you are assigning new values to speed.

# Initialize the speed variable
speed <- 80

# 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 80"
## [1] "Slow down big time!"
## [1] "Your speed is 69"
## [1] "Slow down big time!"
## [1] "Your speed is 58"
## [1] "Slow down big time!"
## [1] "Your speed is 47"
## [1] "Slow down!"
## [1] "Your speed is 41"
## [1] "Slow down!"
## [1] "Your speed is 35"
## [1] "Slow down!"

Stop the while loop: break

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.

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"

Build a while loop from scratch

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.

# Initialize i as 1 
i <- 1

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

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

Loop over a vector

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)
}

## [1] 2
## [1] 3
## [1] 5
## [1] 7
## [1] 11
## [1] 13

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

## [1] 2
## [1] 3
## [1] 5
## [1] 7
## [1] 11
## [1] 13

Write a for loop that iterates over all the elements of linkedin and prints out every element separately.

# 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

Loop over a list

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)
}

## [1] 2
## [1] 3
## [1] 5
## [1] 7
## [1] 11
## [1] 13

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

## [1] 2
## [1] 3
## [1] 5
## [1] 7
## [1] 11
## [1] 13

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.

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 (info in nyc) {
  print(info)
}

## [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

Loop over a matrix

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
  }
}

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.

#create the ttt matrix
R1 <- c("O", NA, "X")
R2 <- c(NA, "O", "O")
R3 <- c("X", NA, "X")

ttt <- rbind(R1, R2, R3)

# 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 O"
## [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 O"
## [1] "On row 2 and column 3 the board contains O"
## [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"

Mix it up with control flow

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.

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!”

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

Next, you break it

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.

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).

#break statement example
x <- 1:5
for (val in x) {
if (val == 3){
break
}
print(val)
}

## [1] 1
## [1] 2

#next statement example
x <- 1:5
for (val in x) {
if (val == 3){
next
}
print(val)
}

## [1] 1
## [1] 2
## [1] 4
## [1] 5

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

# Adapt/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!"

Build a for loop from scratch

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

• 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
  }
  if (char == "u") {
    break
  }
}

# Print out rcount
rcount

## [1] 5

Functions

Functions are an extremely important concept in almost every programming language, and R is no different. Learn what functions are and how to use them—then take charge by writing your own functions.

Function documentation

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)

## starting httpd help server ... done

?sample

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

args(sample)

## function (x, size, replace = FALSE, prob = NULL) 
## NULL

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

?mean

args(mean)

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

Use a function

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.

• 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

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. Notice that both trim and na.rm have default values. This makes them optional arguments.

• 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

When the trim argument is not zero, it chops off a fraction (equal to trim) of the vector you pass as argument x.

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, ...)

• 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

Functions inside functions

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))

## [1] "Your speed is 31"

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

• Use abs() on linkedin - facebook to get the absolute differences between the daily Linkedin and Facebook profile views. Place the call to abs() 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
mean(abs(linkedin - facebook), na.rm = TRUE)

## [1] 4.8

Write your own function

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.

• 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(x) {
    x ^ 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

# Create a function to calculate the area of a circle
area_circle <- function(r) {
  area <- pi * r ^ 2
  return(area)
}

area_circle(7)

## [1] 153.938

# Functions with multiple expressions
circle <- function(r) {
  area <- pi * r ^ 2
  circumference <- 2 * pi * r
  return(list(area = area, circumference = circumference))
}

circle(r = 3)

## $area
## [1] 28.27433
## 
## $circumference
## [1] 18.84956

# Functions work with vectors as well
x <- c(3, 5, 2)

circle <- function(r) {
  area <- pi * r ^ 2
  circumference <- 2 * pi * r
  return(list(area = area, circumference = circumference))
}

circle(x)

## $area
## [1] 28.27433 78.53982 12.56637
## 
## $circumference
## [1] 18.84956 31.41593 12.56637

# The result can be printed as a data.frame
circle <- function(r) {
  area <- pi * r ^ 2
  circumference <- 2 * pi * r
  return(data.frame(area = area, circumference = circumference))
}

circle(x)

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()

## [1] 4

• 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!")
    return(TRUE)
}

# Call the function hello()
hello()

## [1] "Hi there!"

## [1] TRUE

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
}

• 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)
}

pow_two(5)

## [1] "5 to the power two equals 25"

## [1] 25

pow_two(4, FALSE)

## [1] 16

Function scoping

Function scoping implies that variables that are defined inside a function are not accessible outside that function:

pow_two <- function(x) {
  y <- x ^ 2
  return(y)
}
pow_two(4)

## [1] 16

y

## [1] 7

x

## [1] 3 5 2

y was defined inside the pow_two() function and therefore it is not accessible outside of that function. This is also true for the function’s arguments of course - x in this case.

R passes arguments by value

R passes arguments by value. What does this mean? Simply put, it means that an R function cannot change the variable that you input to that function:

triple <- function(x) {
  x <- 3 * x
  x
}
a <- 5
triple(a)

## [1] 15

a

## [1] 5

Inside the triple() function, the argument x gets overwritten with its value times three. Afterwards this new x is returned. If you call this function with a variable a set equal to 5, you obtain 15. But did the value of a change? If R were to pass a to triple() by reference, the override of the x inside the function would ripple through to the variable a, outside the function. However, R passes by value, so the R objects you pass to a function can never change unless you do an explicit assignment. a remains equal to 5, even after calling triple(a).

increment <- function(x, inc = 1) {
  x <- x + inc
  x
}
count <- 5
a <- increment(count, 2)
b <- increment(count)
count <- increment(count, 2)

increment(3)

## [1] 4

increment(count)

## [1] 8

increment(a)

## [1] 8

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.

R you functional?

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.

• 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.

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

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.

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

The apply family

Whenever you’re using a for loop, you may want to revise your code to see whether you can use the lapply function instead. Learn all about this intuitive way of applying a function over a list or a vector, and how to use its variants, sapply and vapply.

Use lapply with a built-in R function

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.

lapply always returns a list, regardless of the class of the input.

x <- list(a = 1:5, b = rnorm(10))
lapply(x, mean)

## $a
## [1] 3
## 
## $b
## [1] -0.04832504

x <- 1:4
lapply(x, runif)

## [[1]]
## [1] 0.3300752
## 
## [[2]]
## [1] 0.1683886 0.4992346
## 
## [[3]]
## [1] 0.2508182 0.8320721 0.3940968
## 
## [[4]]
## [1] 0.18363168 0.98672453 0.69794763 0.07974468

x <- 1:4
lapply(x, runif, min = 0, max = 10)

## [[1]]
## [1] 7.964619
## 
## [[2]]
## [1] 0.877470 9.620131
## 
## [[3]]
## [1] 7.366750 7.530466 5.679713
## 
## [[4]]
## [1] 0.3148979 7.7827947 2.0472940 4.6496357

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!

• 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 = ":")
split_math

## [[1]]
## [1] "GAUSS" "1777" 
## 
## [[2]]
## [1] "BAYES" "1702" 
## 
## [[3]]
## [1] "PASCAL" "1623"  
## 
## [[4]]
## [1] "PEARSON" "1857"

# 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"

Use lapply with your own function

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.

• 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)
names

## [[1]]
## [1] "gauss"
## 
## [[2]]
## [1] "bayes"
## 
## [[3]]
## [1] "pascal"
## 
## [[4]]
## [1] "pearson"

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

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

## [[1]]
## [1] "1777"
## 
## [[2]]
## [1] "1702"
## 
## [[3]]
## [1] "1623"
## 
## [[4]]
## [1] "1857"

lapply and anonymous functions

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 }

## function(x) { 3 * x }

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

## [[1]]
## [1] 3
## 
## [[2]]
## [1] 6
## 
## [[3]]
## [1] 9

• 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.

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

## [[1]]
## [1] "gauss" "1777" 
## 
## [[2]]
## [1] "bayes" "1702" 
## 
## [[3]]
## [1] "pascal" "1623"  
## 
## [[4]]
## [1] "pearson" "1857"

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

## [[1]]
## [1] "gauss"
## 
## [[2]]
## [1] "bayes"
## 
## [[3]]
## [1] "pascal"
## 
## [[4]]
## [1] "pearson"

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

## [[1]]
## [1] "1777"
## 
## [[2]]
## [1] "1702"
## 
## [[3]]
## [1] "1623"
## 
## [[4]]
## [1] "1857"

Use lapply with additional arguments

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)

## [[1]]
## [1] 3
## 
## [[2]]
## [1] 6
## 
## [[3]]
## [1] 9

• 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)

Apply functions that return NULL

In all of the previous exercises, it was assumed that the functions that were applied over vectors and lists actually returned a meaningful result. For example, the tolower() function simply returns the strings with the characters in lowercase. This won’t always be the case. Suppose you want to display the structure of every element of a list. You could use the str() function for this, which returns NULL:

lapply(list(1, "a", TRUE), str)

##  num 1
##  chr "a"
##  logi TRUE

## [[1]]
## NULL
## 
## [[2]]
## NULL
## 
## [[3]]
## NULL

This call actually returns a list, the same size as the input list, containing all NULL values. On the other hand calling str(TRUE) on its own prints only the structure of the logical to the console, not NULL. That’s because str() uses invisible() behind the scenes, which returns an invisible copy of the return value, NULL in this case. This prevents it from being printed when the result of str() is not assigned.

How to use sapply

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, ...)

x <- list(a = 1:4, b = rnorm(10), c = rnorm(20, 1), d = rnorm(100, 5))

sapply(x, mean)

##         a         b         c         d 
## 2.5000000 0.4471128 0.9732262 4.9353707

sapply(list(runif (10), runif (10)), 
       function(x) c(min = min(x), mean = mean(x), max = max(x)))

##            [,1]      [,2]
## min  0.01052101 0.1358788
## mean 0.54630780 0.5175346
## max  0.97577471 0.9104646

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.

# 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

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.

sapply with your own function

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.

• 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.

# 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

sapply with function returning vector

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?

• 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.

# 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

sapply can’t simplify, now what?

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.

# 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

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().

sapply with functions that return NULL

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.

# 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

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.

Use vapply

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.

# 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

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.

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

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.

# 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

From sapply to vapply

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!).

# 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

Utilities

Mastering R programming is not only about understanding its programming concepts. Having a solid understanding of a wide range of R functions is also important. This chapter introduces you to many useful functions for data structure manipulation, regular expressions, and working with times and dates.

Mathematical utilities

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 scientist 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. Calculate the sum of the absolute rounded values of the training errors.

errors <- c(1.9, -2.6, 4.0, -9.5, -3.4, 7.3)

# Sum of absolute rounded values of errors
sum(abs(round(errors)))

## [1] 29

vec1 <- c(1.5, 2.5, 8.4, 3.7, 6.3)
vec2 <- rev(vec1)

# Fix the error
mean(c(abs(vec1), abs(vec2)))

## [1] 4.48

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!

Data Utilities

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.

# 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

# Creating a sequence of numbers
seq(1, 7)

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

seq(1, 10, by = 2)

## [1] 1 3 5 7 9

# Repeating a number 6 times
rep(5, 6)

## [1] 5 5 5 5 5 5

# Combining the two
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

Beat Gauss using R

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?

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.

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

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

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

## [1] 87029

grepl & grep

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.

# 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("edu", emails)

## [1]  TRUE  TRUE FALSE  TRUE  TRUE FALSE

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

# Subset emails using hits
emails[hits]

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

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.

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.

# 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"

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!

sub & gsub

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.

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.

x <- c("AEDFE", "EBGH", "HEAO", "PLETE")

#replace Es with Ks. Notice that sub() only replaces the first occurrences
sub("E", "K", x)

## [1] "AKDFE" "KBGH"  "HKAO"  "PLKTE"

#replace all occurrences with gsub()
x <- c("AEDFE", "EBGH", "HEAO", "PLETE")
gsub("E", "K", x)

## [1] "AKDFK" "KBGH"  "HKAO"  "PLKTK"

# 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"

Notice how only the valid .edu addresses are changed while the other emails remain unchanged.

Right here, right now

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.

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

## [1] "2020-12-16"

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

## [1] 18612

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

## [1] "2020-12-16 12:53:11 PKT"

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

## [1] 1608105191

Create and format dates

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")

## [1] "1982-01-13"

as.Date("Jan-13-82", format = "%b-%d-%y")

## [1] "1982-01-13"

as.Date("13 January, 1982", format = "%d %B, %Y")

## [1] "1982-01-13"

date <- "23rd-May, 2020"
as.Date(date, format = "%drd-%b, %Y")

## [1] "2020-05-23"

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")

## [1] "16 December, 2020"

format(Sys.Date(), format = "Today is a %A!")

## [1] "Today is a Wednesday!"

# 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"

Create and format times

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.

# 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"

## [1] "01"
format(time2, "%I:%M %p")

## [1] "02:23 PM"

Calculations with Dates

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.

today <- Sys.Date()
today + 1

## [1] "2020-12-17"

today - 1

## [1] "2020-12-15"

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

## Time difference of 13 days

• 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 <- "2020-11-28"
day2 <- "2020-11-30"
day3 <- "2020-12-05"
day4 <- "2020-12-11"
day5 <- "2020-12-16"

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

## Time difference of 18 days

# Create vector pizza
pizza <- as.Date(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

Calculations with Times

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

## [1] "2020-12-16 13:53:11 PKT"

now - 3600 * 24     # subtract a day

## [1] "2020-12-15 12:53:11 PKT"

Adding or subtracting 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

## Time difference of 27792.53 days

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.

login <- c("2020-12-02 10:18:04 UTC", "2020-12-07 09:14:18 UTC", "2020-12-07 12:21:51 UTC", "2020-12-07 12:37:24 UTC", "2020-12-09 21:37:55 UTC")
logout <- c("2020-12-02 10:56:29 UTC", "2020-12-07 09:14:52 UTC", "2020-12-07 12:35:48 UTC", "2020-12-07 13:17:22 UTC", "2020-12-09 22:08:47 UTC")

# Calculate the difference between login and logout: time_online
time_online = as.POSIXct(logout) - as.POSIXct(login)

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

Time is of the essence

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)

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")
astro_dates

## [1] "2015-03-20" "2015-06-25" "2015-09-23" "2015-12-22"

# 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