CODEWITHSUNDEEP
rapply
etulf
etulf

Reputation: 3

Apply returns different values depending on which columns are included

If I have a data frame like this:

df <- cbind.data.frame(c("a", "b", "a", "b", "b"), c(1,0,0,1,0), c(0, NA, 0, 0, 1))

What should I do to return 1 for column 3 regardless of whether I've included the character column?

apply(df, 2, FUN = function(x){sum(x == 1 & !is.na(x))})

Returns 0 for column 3

apply(df[,2:3], 2, FUN = function(x){sum(x == 1 & !is.na(x))})

Returns 1 for column 3

Upvotes: 0

Views: 59

Answers (1)

Andre Wildberg
Andre Wildberg

Reputation: 19191

An explanation why apply on the whole data set gives different results compared to the subset (df <> df[,2:3]).

See how apply treats the given data if it's heterogeneous (character and numeric)

apply(df, 2, FUN = function(x) x)
     c("a", "b", "a", "b", "b") c(1, 0, 0, 1, 0) c(0, NA, 0, 0, 1)
[1,] "a"                        "1"              " 0"
[2,] "b"                        "0"              NA
[3,] "a"                        "0"              " 0"
[4,] "b"                        "1"              " 0"
[5,] "b"                        "0"              " 1"

Since

apply(X, MARGIN, ... expects -> X: an array, including a matrix

and it includes the first character column the result gets cast to character (only data.frame and list can hold different data types) and the 3rd column max cell length is 2 because of the NA, all elements get extended to length 2 by padding with space (" 1", which is != 1). There is a workaround using trimws but that's overcomplicating things. Rather

using apply on the homogeneous subset which keeps numeric

apply(df[,2:3], 2, function(x) x)
     c(1, 0, 0, 1, 0) c(0, NA, 0, 0, 1)
[1,]                1                 0
[2,]                0                NA
[3,]                0                 0
[4,]                1                 0
[5,]                0                 1

or use sapply, since we're operating on columns anyways.

Upvotes: 0

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