*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 = .14),
SBP = rnorm(50, mean = 134, sd = 4))
head(dat) Age Weight Height SBP
1 46.75466 91.81194 1.772580 134.8399
2 42.97497 95.55737 1.786904 134.6545
3 38.67239 86.21419 1.528138 133.9183
4 34.06415 73.61920 2.007315 131.3021
5 55.05617 86.58837 1.679040 141.1234
6 43.73024 86.97343 1.580100 141.9637Let’s calculate the mean value of each column:
dat_column_mean <- apply(dat, MARGIN = 2, FUN = mean)
dat_column_mean Age Weight Height SBP
41.010220 80.603914 1.720671 134.606120
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
41.010220 80.603914 1.720671 134.606120 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 96.17808 92.66500 89.10370 86.93418
2 8002 108.34753 103.64442 100.28809 98.04957
3 8003 98.75378 95.62569 91.90074 88.87626
4 8004 110.49883 107.35605 105.33629 102.29831
5 8005 111.12856 110.13624 106.59963 104.82050
6 8006 114.77758 114.66860 112.07814 109.79900
7 8007 110.20985 108.97289 105.40417 103.43360
8 8008 96.68164 93.65599 89.78490 88.79433
9 8009 98.57767 96.06798 92.91079 90.39733
10 8010 124.40278 123.05070 120.77419 118.35642
11 8011 112.60340 111.00301 107.96205 105.58551
12 8012 109.19388 106.74415 104.64558 102.66538
13 8013 98.61217 96.75388 93.55110 91.18758
14 8014 115.31654 111.60113 108.38239 104.85049
15 8015 124.17488 122.16698 119.99058 119.66895
16 8016 98.61493 97.69010 95.03227 93.09582
17 8017 101.64022 98.44237 94.08062 93.46985
18 8018 128.12569 125.73114 123.66003 122.26057
19 8019 107.36327 103.56979 101.89622 100.90789
20 8020 115.04972 112.47232 108.60742 105.74453Let’s get the mean weight per week:
apply(dat2[, -1], 2, mean)Weight_week_1 Weight_week_3 Weight_week_5 Weight_week_7
109.0126 106.6009 103.5994 101.5598 Let’s get the mean weight per individual across all weeks:
apply(dat2[, -1], 1, mean) [1] 91.22024 102.58240 93.78912 106.37237 108.17123 112.83083 107.00513
[8] 92.22922 94.48844 121.64602 109.28849 105.81225 95.02618 110.03764
[15] 121.50035 96.10828 96.90826 124.94436 103.43429 110.46850
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] 39.68335
$Weight
[1] 79.52706
$Height
[1] 1.717384
$SBP
[1] 134.1549To 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
39.683351 79.527060 1.717384 134.154949 dat_summary <- data.frame(Mean = sapply(dat, mean),
SD = sapply(dat, sd))
dat_summary Mean SD
Age 41.010220 7.6629240
Weight 80.603914 9.8307274
Height 1.720671 0.1330028
SBP 134.606120 3.680784924.3.1 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 96.17808 92.66500 89.10370 86.93418
2 8002 108.34753 103.64442 100.28809 98.04957
3 8003 98.75378 95.62569 91.90074 88.87626
4 8004 110.49883 107.35605 105.33629 102.29831
5 8005 111.12856 110.13624 106.59963 104.82050
6 8006 114.77758 114.66860 112.07814 109.79900logical index of numeric columns:
numidl <- sapply(dat2, is.numeric)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 = .1) Age Weight Height SBP
39.683351 79.527060 1.717384 134.154949 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,] 23.49001 58.87642 1.467072 127.0328
[2,] 60.51084 112.78688 2.007315 141.9637If 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
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:
mean_Age_by_Group <- tapply(dat[["Age"]], dat["Group"], mean)
mean_Age_by_GroupGroup
A B C
40.06771 42.89581 40.01759 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 46.75466 91.81194 1.772580 134.8399
2 42.97497 95.55737 1.786904 134.6545
3 38.67239 86.21419 1.528138 133.9183
4 34.06415 73.61920 2.007315 131.3021
5 55.05617 86.58837 1.679040 141.1234
6 43.73024 86.97343 1.580100 141.9637 Age Weight Height SBP
[1,] 2185.998 8429.432 3.142041 18181.81
[2,] 1846.848 9131.211 3.193027 18131.84
[3,] 1495.553 7432.887 2.335205 17934.12
[4,] 1160.366 5419.787 4.029315 17240.24
[5,] 3031.182 7497.546 2.819176 19915.81
[6,] 1912.334 7564.378 2.496718 20153.69Let’s do the same as above, but this time using an anonymous function:
Age Weight Height SBP
[1,] 2185.998 8429.432 3.142041 18181.81
[2,] 1846.848 9131.211 3.193027 18131.84
[3,] 1495.553 7432.887 2.335205 17934.12
[4,] 1160.366 5419.787 4.029315 17240.24
[5,] 3031.182 7497.546 2.819176 19915.81
[6,] 1912.334 7564.378 2.496718 20153.69The 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
41.010220 80.603914 1.720671 134.606120 NA [1] 41.010220 80.603914 1.720671 134.606120# 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.