*apply() function family summary (Best to read through this chapter first and then refer back to this figure)20 Loop Functions
Loop functions are some of the most widely used R functions. They replace longer expressions created with a for loop, for example.
They can result in more compact and readable code.
| Function | Description |
|---|---|
apply() |
Apply function over array margins (i.e. over one or more dimensions) |
lapply() |
Return a list where each element is the result of applying a function to each element of the input |
sapply() |
Same as lapply(), but returns the simplest possible R object (instead of always returning a list) |
vapply() |
Same as sapply(), but with a pre-specified return type: this is safer and may also be faster |
tapply() |
Apply a function to elements of groups defined by a factor |
mapply() |
Multivariate sapply(): Apply a function using the 1st elements of the inputs vectors, then using the 2nd, 3rd, etc. |
20.1 apply()
Tip
apply() applies a function over one or more dimensions of an array of 2 dimensions or more (this includes matrices) or a data frame:
apply(array, MARGIN, FUN)
MARGIN can be an integer vector or character indicating the dimensions over which ‘FUN’ will be applied.
By convention, rows come first (just like in indexing), therefore:
-
MARGIN = 1: apply function on each row -
MARGIN = 2: apply function on each column
Let’s create an example dataset:
dat <- data.frame(Age = rnorm(50, mean = 42, sd = 8),
Weight = rnorm(50, mean = 80, sd = 10),
Height = rnorm(50, mean = 1.72, sd = .14),
SBP = rnorm(50, mean = 134, sd = 4))
head(dat) Age Weight Height SBP
1 32.53260 73.49671 1.709358 132.2705
2 46.06687 77.60429 1.633672 135.0700
3 31.83870 94.02186 1.703996 124.1504
4 39.95082 94.08280 1.664703 134.8827
5 28.54416 76.92390 1.887825 138.6463
6 51.04894 80.77547 1.904373 128.8690Let’s calculate the mean value of each column:
dat_column_mean <- apply(dat, MARGIN = 2, FUN = mean)
dat_column_mean Age Weight Height SBP
41.929840 81.348768 1.725531 133.625099
Tip
Hint: It is possibly easiest to think of the “MARGIN” as the dimension you want to keep.
In the above case, we want the mean for each variable, i.e. we want to keep columns and collapse rows.
Purely as an example to understand what apply() does, here is the equivalent procedure using a for-loop. You notice how much more code is needed, and why apply() and similar functions might be very convenient for many different tasks.
dat_column_mean <- numeric(ncol(dat))
names(dat_column_mean) <- names(dat)
for (i in seq(dat)) {
dat_column_mean[i] <- mean(dat[, i])
}
dat_column_mean Age Weight Height SBP
41.929840 81.348768 1.725531 133.625099 Let’s create a different example dataset, where we record weight at multiple timepoints:
dat2 <- data.frame(ID = seq(8001, 8020),
Weight_week_1 = rnorm(20, mean = 110, sd = 10))
dat2$Weight_week_3 <- dat2$Weight_week_1 + rnorm(20, mean = -2, sd = 1)
dat2$Weight_week_5 <- dat2$Weight_week_3 + rnorm(20, mean = -3, sd = 1.1)
dat2$Weight_week_7 <- dat2$Weight_week_5 + rnorm(20, mean = -1.8, sd = 1.3)
dat2 ID Weight_week_1 Weight_week_3 Weight_week_5 Weight_week_7
1 8001 109.46529 107.68752 103.18683 99.76605
2 8002 107.81770 106.65251 102.71864 99.22260
3 8003 106.52477 104.02491 100.86989 99.44099
4 8004 109.40128 109.87463 105.56668 103.01342
5 8005 101.07879 99.18237 94.85858 94.67359
6 8006 109.82664 107.51991 105.16933 99.25406
7 8007 112.12645 109.20799 106.24329 103.40150
8 8008 114.35938 112.30606 110.49744 110.15284
9 8009 111.84084 110.56277 106.98929 105.50829
10 8010 105.25903 102.51100 99.08774 99.11549
11 8011 105.92403 102.51770 99.00781 94.38968
12 8012 108.41214 107.45216 104.53650 105.10564
13 8013 100.94935 99.49458 95.86894 93.88306
14 8014 107.00289 103.60279 100.81051 97.97840
15 8015 99.72552 99.61076 96.09652 95.33871
16 8016 109.63041 107.45409 103.77169 103.16132
17 8017 122.96651 121.40089 118.50415 117.71091
18 8018 124.99564 121.28400 118.39624 116.35645
19 8019 99.51919 98.44971 94.45205 91.63567
20 8020 116.86688 114.68809 111.89008 107.90306Let’s get the mean weight per week:
apply(dat2[, -1], 2, mean)Weight_week_1 Weight_week_3 Weight_week_5 Weight_week_7
109.1846 107.2742 103.9261 101.8506 Let’s get the mean weight per individual across all weeks:
apply(dat2[, -1], 1, mean) [1] 105.02642 104.10286 102.71514 106.96400 97.44833 105.44249 107.74481
[8] 111.82893 108.72530 101.49332 100.45981 106.37661 97.54898 102.34865
[15] 97.69288 106.00438 120.14562 120.25808 96.01415 112.83703
Caution
apply() converts 2-dimensional objects to matrices before applying the function. Therefore, if applied on a data.frame with mixed data types, it will be coerced to a character matrix.
This is explained in the apply() documentation under “Details”:
“If X is not an array but an object of a class with a non-null dim value (such as a data frame), apply attempts to coerce it to an array via as.matrix if it is two-dimensional (e.g., a data frame) or via as.array.”
Because of the above, see what happens when you use apply on the iris data.frame which contains 4 numeric variables and one factor:
str(iris)'data.frame': 150 obs. of 5 variables:
$ Sepal.Length: num 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...
$ Sepal.Width : num 3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ...
$ Petal.Length: num 1.4 1.4 1.3 1.5 1.4 1.7 1.4 1.5 1.4 1.5 ...
$ Petal.Width : num 0.2 0.2 0.2 0.2 0.2 0.4 0.3 0.2 0.2 0.1 ...
$ Species : Factor w/ 3 levels "setosa","versicolor",..: 1 1 1 1 1 1 1 1 1 1 ...apply(iris, 2, class)Sepal.Length Sepal.Width Petal.Length Petal.Width Species
"character" "character" "character" "character" "character"
20.2 lapply()
Tip
lapply() applies a function on each element of its input and returns a list of the outputs.
Note: The ‘elements’ of a data frame are its columns (remember, a data frame is a list with equal-length elements). The ‘elements’ of a matrix are each cell one by one, by column. Therefore, unlike apply(), lapply() has a very different effect on a data frame and a matrix. lapply() is commonly used to iterate over the columns of a data frame.
Tip
lapply() is the only function of the *apply() family that always returns a list.
dat_median <- lapply(dat, median)
dat_median$Age
[1] 43.31462
$Weight
[1] 80.78033
$Height
[1] 1.70545
$SBP
[1] 133.9186To understand what lapply() does, here is the equivalent for-loop:
20.3 sapply()
sapply() is an alias for lapply(), followed by a call to simplify2array().
(Check the source code for sapply() by typing sapply at the console).
Note
dat_median <- sapply(dat, median)
dat_median Age Weight Height SBP
43.31462 80.78033 1.70545 133.91862 dat_summary <- data.frame(Mean = sapply(dat, mean),
SD = sapply(dat, sd))
dat_summary Mean SD
Age 41.929840 7.2477392
Weight 81.348768 10.6952567
Height 1.725531 0.1649386
SBP 133.625099 4.061002220.3.1 Example: Get index of numeric variables
Let’s use sapply() to get an index of numeric columns in dat2:
head(dat2) ID Weight_week_1 Weight_week_3 Weight_week_5 Weight_week_7
1 8001 109.4653 107.68752 103.18683 99.76605
2 8002 107.8177 106.65251 102.71864 99.22260
3 8003 106.5248 104.02491 100.86989 99.44099
4 8004 109.4013 109.87463 105.56668 103.01342
5 8005 101.0788 99.18237 94.85858 94.67359
6 8006 109.8266 107.51991 105.16933 99.25406logical index of numeric columns:
numidl <- sapply(dat2, is.numeric)integer index of numeric columns:
20.4 vapply()
Much less commonly used (possibly underused) than lapply() or sapply(), vapply() allows you to specify what the expected output looks like - for example a numeric vector of length 2, a character vector of length 1.
This can have two advantages:
- It is safer against errors
- It will sometimes be a little faster
You add the argument FUN.VALUE which must be of the correct type and length of the expected result of each iteration.
vapply(dat, median, FUN.VALUE = .1) Age Weight Height SBP
43.31462 80.78033 1.70545 133.91862 Here, each iteration returns the median of each column, i.e. a numeric vector of length 1.
Therefore FUN.VALUE can be any numeric scalar.
For example, if we instead returned the range of each column, FUN.VALUE should be a numeric vector of length 2:
Age Weight Height SBP
[1,] 21.73529 53.47993 1.437573 124.1504
[2,] 53.78449 103.32512 2.141495 140.5375If FUN.VALUE does not match the returned value, we get an informative error:
vapply(dat, range, FUN.VALUE = .1)Error in vapply(dat, range, FUN.VALUE = 0.1): values must be length 1,
but FUN(X[[1]]) result is length 2
20.5 tapply()
tapply() is one way (of many) to apply a function on subgroups of data as defined by one or more factors.
In the following example, we calculate the mean Sepal.Length by species on the iris dataset:
mean_Age_by_Group <- tapply(dat[["Age"]], dat["Group"], mean)
mean_Age_by_GroupGroup
A B C
42.38384 39.40224 43.38493 The for-loop equivalent of the above is:
20.6 mapply()
The functions we have looked at so far work well when you iterating over elements of a single object.
mapply() allows you to execute a function that accepts two or more inputs, say fn(x, z) using the i-th element of each input, and will return:fn(x[1], z[1]), fn(x[2], z[2]), …, fn(x[n], z[n])
Let’s create a simple function that accepts two numeric arguments, and two vectors length 5 each:
raise <- function(x, power) x^power
x <- 2:6
p <- 6:2Use mapply to raise each x to the corresponding p:
out <- mapply(raise, x, p)
out[1] 64 243 256 125 36The above is equivalent to:
20.7 *apply()ing on matrices vs. data frames
To consolidate some of what was learned above, let’s focus on the difference between working on a matrix vs. a data frame.
First, let’s create a matrix and a data frame with the same data:
Feature_1 Feature_2 Feature_3 Feature_4 Feature_5
[1,] 21 31 41 51 61
[2,] 22 32 42 52 62
[3,] 23 33 43 53 63
[4,] 24 34 44 54 64
[5,] 25 35 45 55 65
[6,] 26 36 46 56 66
[7,] 27 37 47 57 67
[8,] 28 38 48 58 68
[9,] 29 39 49 59 69
[10,] 30 40 50 60 70adf <- as.data.frame(amat)
adf Feature_1 Feature_2 Feature_3 Feature_4 Feature_5
1 21 31 41 51 61
2 22 32 42 52 62
3 23 33 43 53 63
4 24 34 44 54 64
5 25 35 45 55 65
6 26 36 46 56 66
7 27 37 47 57 67
8 28 38 48 58 68
9 29 39 49 59 69
10 30 40 50 60 70We’ve seen that with apply() we specify the dimension to operate on and it works the same way on both matrices and data frames:
apply(amat, 2, mean)Feature_1 Feature_2 Feature_3 Feature_4 Feature_5
25.5 35.5 45.5 55.5 65.5 apply(adf, 2, mean)Feature_1 Feature_2 Feature_3 Feature_4 Feature_5
25.5 35.5 45.5 55.5 65.5 However, sapply() (and lapply(), vapply()) acts on each element of the object, therefore it is not meaningful to pass a matrix to it:
sapply(amat, mean) [1] 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45
[26] 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70The above returns the mean of each element, i.e. the element itself, which is meaningless.
Since a data frame is a list, and its columns are its elements, it works great for column operations on data frames:
sapply(adf, mean)Feature_1 Feature_2 Feature_3 Feature_4 Feature_5
25.5 35.5 45.5 55.5 65.5 If you want to use sapply() on a matrix, you could iterate over an integer sequence as shown in the previous section:
This is shown to help emphasize the differences between the function and the data structures. In practice, you would use apply() on a matrix.
20.8 Anonymous functions
Anonymous functions are just like regular functions but they are not assigned to an object - i.e. they are not “named”.
They are usually passed as arguments to other functions to be used once, hence no need to assign them.
Anonymous functions are often used with the apply family of functions.
Example of a simple regular function:
squared <- function(x) {
x^2
}Since this is a short function definition, it can also be written in a single line:
squared <- function(x) x^2An anonymous function definition is just like a regular function - minus it is not assigned:
function(x) x^2Since R version 4.1 (May 2021), a compact anonymous function syntax is available, where a single back slash replaces function:
\(x) x^2Let’s use the squared() function within sapply() to square the first four columns of the iris dataset. In these examples, we often wrap functions around head() which prints the first few lines of an object to avoid:
head(dat[, 1:4]) Age Weight Height SBP
1 32.53260 73.49671 1.709358 132.2705
2 46.06687 77.60429 1.633672 135.0700
3 31.83870 94.02186 1.703996 124.1504
4 39.95082 94.08280 1.664703 134.8827
5 28.54416 76.92390 1.887825 138.6463
6 51.04894 80.77547 1.904373 128.8690 Age Weight Height SBP
[1,] 1058.3700 5401.766 2.921904 17495.49
[2,] 2122.1567 6022.426 2.668883 18243.89
[3,] 1013.7028 8840.110 2.903603 15413.31
[4,] 1596.0681 8851.573 2.771235 18193.33
[5,] 814.7693 5917.286 3.563883 19222.79
[6,] 2605.9941 6524.677 3.626636 16607.22Let’s do the same as above, but this time using an anonymous function:
Age Weight Height SBP
[1,] 1058.3700 5401.766 2.921904 17495.49
[2,] 2122.1567 6022.426 2.668883 18243.89
[3,] 1013.7028 8840.110 2.903603 15413.31
[4,] 1596.0681 8851.573 2.771235 18193.33
[5,] 814.7693 5917.286 3.563883 19222.79
[6,] 2605.9941 6524.677 3.626636 16607.22The entire anonymous function definition is passed to the FUN argument.
20.9 Iterating over a sequence instead of an object
With lapply(), sapply() and vapply() there is a very simple trick that may often come in handy:
Instead of iterating over elements of an object, you can iterate over an integer index of whichever elements you want to access and use it accordingly within the anonymous function.
This alternative approach is much closer to how we would use an integer sequence in a for loop.
It will be clearer through an example, where we get the mean of the first four columns of iris:
Warning in mean.default(i): argument is not numeric or logical: returning NA Age Weight Height SBP Group
41.929840 81.348768 1.725531 133.625099 NA [1] 41.929840 81.348768 1.725531 133.625099# equivalent to:
for (i in 1:4) {
mean(dat[, i])
}Notice that in this approach, since you are not passing the object (dat, in the above example) as the input to lapply(), it needs to be accessed within the anonymous function.