Conditional Structure in R

Conditionals and control flow

To be TRUE or not be TRUE, that’s the question. In this chapter you’ll learn about relational operators to see how R objects compare and logical operators to combine logicals. Next, 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 (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

Greater and less than

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

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

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.

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

Wonderful! Have a look at the console output. Your LinkedIn profile was pretty popular on the sixth day, but less so on the fourth and fifth day.

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.

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

& 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.

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

Great! Have one last look at the console before proceeding; do the results of the different expressions make sense?

& 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?

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 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

Bravo! 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. Can you draw similar conclusions for the other results?

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))

Possible answers:

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

Great!

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.

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.

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

li_df <- 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)

li_df <- matrix(li_df, nrow = 50, byrow = TRUE)

colnames(li_df) <- c("day1", "day2", "day3", "day4", "day5", "day6", "day7")

li_df <- data.frame(li_df)

# Select the second column, named day2, from li_df: second
second <- li_df[, 2]

# 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[, 2] > 25 | li_df[, 2] < 5)

## [1] 16

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.

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!")
}

Great! Try to see what happens if you change the medium and num_views variables and run your code again. Let’s further customize these if statements in the next exercise.

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!

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!"

Great job! You also had Facebook information available, remember? Time to add some more statements to our control structures using else if!

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!

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.

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:

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

Have a look at the following statements:

If number is set to 6, “small” gets printed to the console.
If number is set to 100, R prints out “medium”.
If number is set to 4, “extra small” gets printed out to the console.
If number is set to 2500, R will generate an error, as result will not be defined.

Select the option that lists all the true statements.

Possible answers:

(2) and (4)
(1) and (4)
(1) and (3)
(2) and (3)

Wonderful! If you got this one right, the next exercise will be a piece of cake.

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.

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 <- li + fb
}

# Print the resulting sms to the console
sms

## [1] 24

Conditional Structure in R

Dr. Zulfiqar Ali

Conditionals and control flow

Equality

Instructions

Greater and less than

Instructions

Compare vectors

Instructions

Compare matrices

Instructions

& and |

Instructions

& and | (2)

Instructions

Reverse the result: !

Blend it all together

Instructions

The if statement

Instructions

Add an else

Instructions

Customize further: else if

Instructions

Else if 2.0

Take control!

Instructions