Apply Functions


In R, we might encounter situations where we need to iterate over several (or all) of the elements in a data frame. With languages like Python or Java, for-loops and while-loops are indispensable when dealing with iteration, but in R, we have the apply functions, which offer a faster, more concise way of applying a function to a data set.

There’s a great online encyclopedia on R that inspired and informed many examples for this page. Check them out!


The basic apply function takes an array/matrix, a MARGIN parameter for the desired dimension, and the function you want to apply.

Our test matrix will be 5x5, where the numbers 1-25 will populate each column sequentially:

cubed5 <- matrix(c(1:5, 6:10, 11:15, 16:20, 21:25), nrow = 5, ncol = 5)
     [,1] [,2] [,3] [,4] [,5]
[1,]    1    6   11   16   21
[2,]    2    7   12   17   22
[3,]    3    8   13   18   23
[4,]    4    9   14   19   24
[5,]    5   10   15   20   25
  • insert examples for apply here

Ultimately, that’s all there is to apply. One important note is that apply will not work for vectors — we discuss vector apply functions in a moment.

squares <- c(1, 4, 9, 16, 25)
apply(squares, 1, function(x) x ^ 0.5)
`Error in apply(squares, 1, function(x) x^0.5) :
  dim(X) must have a positive length`

lapply & sapply

lapply applies a function to each element of a list, then returns a list that’s been altered by the function. Since there is only one dimension in a list, the MARGIN parameter does not apply. Let’s use sum on the squares vector from before:

lapply(squares, sum)
[1] 1

[1] 4

[1] 9

[1] 16

[1] 25

This is obviously not what we wanted — the problem here is that vectors and lists are not the same thing and lapply treats every element in the vector like its own list. Casting squares to a list will fix the issue:

lapply(list(squares), sum)
[1] 55

sapply will function identically to lapply unless the output can be simplified, in which case sapply executes that simplification. The following occurs when we run sapply in place of lapply on our squares vector.

sapply(squares, sum)
[1]  1  4  9 16 25

Unless you explicitly need a list, sapply can often be the more advantageous function because of its output simplification.

How do I get the mean values of the list pamphlets with attributes pages, words, and letters?

lapply(my_list, mean)
[1] 3

[1] 30

[1] 300


The documentation definition for tapply is a bit more specific than the others, where the arguments are now (X, INDEX, FUN), with X being an object where the split function applies, INDEX is a factor by which X is grouped, and FUN is function as before.

To simplify this definition, we can say tapply applies FUN to X when X is grouped by INDEX. Consider the following: we have a grades data.frame that contains information for grade, year, and sex for several students. We can use tapply to get the average grade by year in a simple way.

   grade      year  sex
1    100    junior    M
2     99 sophomore    F
3     75 sophomore    M
4     74 sophomore    M
5     44    senior    F
6     69    junior    M
7     88    junior    F
8     99    senior <NA>
9     90  freshman    M
10    92    junior    F

The solution begins below.

tapply(grades$grade, grades$year, mean)
 freshman    junior    senior sophomore
 90.00000  87.25000  71.50000  82.66667

We can use the optional arguments here to remove any rows that contain missing data.

tapply(grades$grade, grades$year, mean, na.rm=T)
##  freshman    junior    senior sophomore
##  90.00000  87.25000  44.00000  82.66667


How can I find the average of several variables in the flight dataset using 1 line of lapply code?

We can store the data for 2003 flights as follows:

myDF <- read.csv("/depot/datamine/data/flights/subset/2003.csv")

We can categorize the flight distances in groups of <100 miles, 100-200 miles, 200-500 miles, 500-1000 miles, 1000-2000 miles, and 2000+ miles using the cut function, then tabulating it

my_distance_categories <- cut(myDF$Distance, breaks = c(0,100,200,500,1000,2000,Inf), include.lowest=T)

We can get the averages of all applicable flights for 4 variables, broken down by the distance categories we just defined.

tapply(myDF$DepDelay, my_distance_categories, mean, na.rm=T)  # the DepDelay in each category
tapply(myDF$ArrDelay, my_distance_categories, mean, na.rm=T)  # the ArrDelay in each category
tapply(myDF$TaxiOut, my_distance_categories, mean, na.rm=T)  # the time to TaxiOut in each category
tapply(myDF$TaxiIn, my_distance_categories, mean, na.rm=T)  # the time to TaxiIn in each category

However, we can condense this to one line using lapply according to the prompt. To make it easier to read, we can make a temporary data frame flights_by_distance with these 4 variables. Then we split the data into 6 data.frames using the distance categories, yielding averages for DepDelay, ArrDelay, TaxiOut, and TaxiIn. This will agree exactly with the results of the 4 separate tapply functions, but it only takes us 1 call to lapply!

flights_by_distance <- split( data.frame(myDF$DepDelay, myDF$ArrDelay, myDF$TaxiOut, myDF$TaxiIn), my_distance_categories )
lapply( flights_by_distance, colMeans, na.rm=T )

How can I find the average of variables DRUNK_DR, FATALS, and PERSONS in the fars dataset using 1 line of lapply code?

This is a question that was asked in previous STAT19000 classes when the apply functions are introduced. We’ll start by reading in the dataset and adding state names.

There are more efficient ways to add the names, but this code mirrors the solution to the previous implementation of this question, which we’ll follow from here on out.

dat <- read.csv("/depot/datamine/data/fars/7581.csv")
state_names <- read.csv("/depot/datamine/data/fars/states.csv")
v <- state_names$state
names(v) <- state_names$code
dat$mystates <- v[as.character(dat$STATE)]

If we wanted to get the averages for the 3 variables in question, we can use tapply independently:

tapply(dat$DRUNK_DR, dat$mystates, mean)
tapply(dat$FATALS, dat$mystates, mean)
tapply(dat$PERSONS, dat$mystates, mean)

However, there is an easier way that also fits the requirements of the prompt. We’ll create the data.frame accidents_by_state with only these 3 variables for readability:

accidents_by_state <- split( data.frame(dat$DRUNK_DR, dat$FATALS, dat$PERSONS), dat$mystates )
lapply( accidents_by_state, colMeans )

The split function creates 51 different data.frames based on the values in mystates, where lapply then uses colMeans as its function to get the averages for our 3 variables. Awesome!

Use the provided code to create a new column transformed in the data.frame example_df. transformed should contain TRUE if the value in column pre_transformed is "t", FALSE if it is "f", and NA otherwise.

string_to_bool <- function(value) {
  if (value == "t") {
  } else if (value == "f") {
  } else {

example_df <- data.frame(pre_transformed=c("f", "f", "t", "f", "something", "t", "else", ""), other=c(1,2,3,4,5,6,7,8))

The solution begins below.

example_df$transformed <- sapply(example_df$pre_transformed, string_to_bool)
  pre_transformed other transformed
1               f     1       FALSE
2               f     2       FALSE
3               t     3        TRUE
4               f     4       FALSE
5       something     5          NA
6               t     6        TRUE
7            else     7          NA
8                     8          NA

Here we have not a question, but a demonstration. We use tapply in various ways on the Amazon Fine Food Reviews dataset.

The goal of our demonstration is to show the most consistently helpful users in this dataset. This is calculated using the HelpfulnessNumerator and HelpfulnessDenominator fields in the dataset. As an example, we find the user that wrote the most reviews.

myDF <- read.csv("/depot/datamine/data/amazon/amazon_fine_food_reviews.csv")

The user in question is A3OXHLG6DIBRW8, which will be further referred to as A3O. The code below provides two summations: the HelpfulnessDenominator sum is the total number of people who read A3O’s reviews, while the HelpfulnessNumerator is the number of people who found their reviews helpful. We can call the sum functions on both, then taking the quotient to get A3O’s Helpfulness proportion.

sum(myDF$HelpfulnessNumerator[myDF$UserId == "A3OXHLG6DIBRW8"])/sum(myDF$HelpfulnessDenominator[myDF$UserId == "A3OXHLG6DIBRW8"])

Instead of grabbing each user individually, we can use tapply to calculate these proportions for all users.

tapply(myDF$HelpfulnessNumerator, myDF$UserId, sum)/tapply(myDF$HelpfulnessDenominator, myDF$UserId, sum)