*apply() function family summary (Best to read through this chapter first and then refer back to this figure)
24 The Apply Family
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. |
24.1 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 = 0.14),
SBP = rnorm(50, mean = 134, sd = 4))
head(dat) Age Weight Height SBP
1 48.16460 88.55235 1.582764 133.7574
2 48.44521 76.57719 1.572241 134.1560
3 33.93637 93.27066 1.671029 137.0709
4 40.44267 93.60998 1.843725 133.3902
5 45.33717 94.70261 1.748060 131.2815
6 39.51068 91.54904 1.623145 131.7906Let’s calculate the mean value of each column:
dat_column_mean <- apply(dat, MARGIN = 2, FUN = mean)
dat_column_mean Age Weight Height SBP
41.889752 83.543067 1.704655 134.091223
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
41.889752 83.543067 1.704655 134.091223 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 105.82478 103.37164 100.71798 99.47703
2 8002 105.48415 100.93332 97.23739 94.26977
3 8003 120.95794 116.98895 114.23984 114.08036
4 8004 114.05294 111.62552 108.93863 107.20867
5 8005 85.67305 82.95501 81.02380 81.03016
6 8006 112.02879 108.79129 103.79578 102.33458
7 8007 99.73759 97.92762 95.96941 92.69766
8 8008 107.44467 104.79624 100.90447 102.27286
9 8009 100.00119 98.00724 95.00727 92.86830
10 8010 97.04331 96.26564 94.54846 93.35433
11 8011 105.20776 102.98488 98.31741 93.60206
12 8012 105.52480 104.96172 103.10312 102.21607
13 8013 116.46133 115.41325 113.64038 110.12251
14 8014 108.69661 108.03059 106.19866 105.26812
15 8015 117.66636 114.88695 111.69464 110.95197
16 8016 109.84529 109.30767 106.00337 102.95822
17 8017 113.64041 111.78063 107.56287 105.95331
18 8018 103.61945 101.57644 96.61125 93.42433
19 8019 126.22185 124.74886 121.80256 119.24130
20 8020 91.46144 91.25641 86.22691 83.97778Let’s get the mean weight per week:
apply(dat2[, -1], 2, mean)Weight_week_1 Weight_week_3 Weight_week_5 Weight_week_7
107.3297 105.3305 102.1772 100.3655 Let’s get the mean weight per individual across all weeks:
apply(dat2[, -1], 1, mean) [1] 102.34786 99.48116 116.56677 110.45644 82.67050 106.73761 96.58307
[8] 103.85456 96.47100 95.30294 100.02803 103.95143 113.90937 107.04850
[15] 113.79998 107.02864 109.73431 98.80787 123.00364 88.23063
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"
24.2 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] 40.45594
$Weight
[1] 83.04049
$Height
[1] 1.67579
$SBP
[1] 133.771To understand what lapply() does, here is the equivalent for-loop:
24.3 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
40.45594 83.04049 1.67579 133.77104 dat_summary <- data.frame(Mean = sapply(dat, mean),
SD = sapply(dat, sd))
dat_summary Mean SD
Age 41.889752 8.0172308
Weight 83.543067 9.1753281
Height 1.704655 0.1356778
SBP 134.091223 4.322413624.3.1 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 105.82478 103.37164 100.71798 99.47703
2 8002 105.48415 100.93332 97.23739 94.26977
3 8003 120.95794 116.98895 114.23984 114.08036
4 8004 114.05294 111.62552 108.93863 107.20867
5 8005 85.67305 82.95501 81.02380 81.03016
6 8006 112.02879 108.79129 103.79578 102.33458logical index of numeric columns:
numidl <- sapply(dat2, is.numeric)
numidl ID Weight_week_1 Weight_week_3 Weight_week_5 Weight_week_7
TRUE TRUE TRUE TRUE TRUE integer index of numeric columns:
24.4 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 = 0.0) Age Weight Height SBP
40.45594 83.04049 1.67579 133.77104 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,] 24.25842 66.66278 1.399964 124.8212
[2,] 63.62827 105.93721 2.081870 146.2187If FUN.VALUE does not match the returned value, we get an informative error:
vapply(dat, range, FUN.VALUE = 0.0)Error in vapply(dat, range, FUN.VALUE = 0): values must be length 1,
but FUN(X[[1]]) result is length 2
24.5 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:
Age Weight Height SBP Group
1 48.16460 88.55235 1.582764 133.7574 A
2 48.44521 76.57719 1.572241 134.1560 C
3 33.93637 93.27066 1.671029 137.0709 B
4 40.44267 93.60998 1.843725 133.3902 B
5 45.33717 94.70261 1.748060 131.2815 B
6 39.51068 91.54904 1.623145 131.7906 Amean_Age_by_Group <- tapply(dat[["Age"]], dat["Group"], mean)
mean_Age_by_GroupGroup
A B C
41.52139 45.20608 38.57328 The for-loop equivalent of the above is:
24.6 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:
24.7 *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.
24.8 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 48.16460 88.55235 1.582764 133.7574
2 48.44521 76.57719 1.572241 134.1560
3 33.93637 93.27066 1.671029 137.0709
4 40.44267 93.60998 1.843725 133.3902
5 45.33717 94.70261 1.748060 131.2815
6 39.51068 91.54904 1.623145 131.7906 Age Weight Height SBP
[1,] 2319.828 7841.518 2.505141 17891.04
[2,] 2346.938 5864.066 2.471943 17997.84
[3,] 1151.677 8699.417 2.792338 18788.42
[4,] 1635.610 8762.828 3.399320 17792.94
[5,] 2055.459 8968.584 3.055714 17234.84
[6,] 1561.094 8381.226 2.634601 17368.77Let’s do the same as above, but this time using an anonymous function:
Age Weight Height SBP
[1,] 2319.828 7841.518 2.505141 17891.04
[2,] 2346.938 5864.066 2.471943 17997.84
[3,] 1151.677 8699.417 2.792338 18788.42
[4,] 1635.610 8762.828 3.399320 17792.94
[5,] 2055.459 8968.584 3.055714 17234.84
[6,] 1561.094 8381.226 2.634601 17368.77The entire anonymous function definition is passed to the FUN argument.
24.9 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
41.889752 83.543067 1.704655 134.091223 NA [1] 41.889752 83.543067 1.704655 134.091223# 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.