I am trying to make a new column in my dataset give a single output for each and every row, depending on the inputs from pre-existing columns.
In this output column, I desire "NA" if any of the input vales in a given row are "0". Otherwise (if none of the inputs are 0), I want the output for that row to be the number of unique values of the inputs.
I thought that the solution would use an ifelse function nested within an apply function, but I get an error that I do not understand.
data$output <- apply(data, 1, function(x) {ifelse(x == 0, NA, length(unique(x)))})
Error in $<-.data.frame(*tmp*, "output", value = c(3L, 3L, 3L, 3L, :
replacement has 3 rows, data has 4
I do not know why the replacement has 3 rows, as I thought apply just does the same function to each of my 4 rows.
You want to check if any of the variables in a row are 0, so you need to use any(x==0) instead of x == 0 in the ifelse statement:
apply(data, 1, function(x) {ifelse(any(x == 0), NA, length(unique(x)))})
# [1] 1 NA 2
Basically ifelse returns a vector of length n if its first argument is of length n. You want one value per row, but are passing more than one with x==0 (the number of values you're passing is equal to the number of columns in your data frame).
Data:
(data <- data.frame(a=c(1, 2, 3), b=c(1, 0, 1)))
# a b
# 1 1 1
# 2 2 0
# 3 3 1
Let n = length(x). ifelse will return rep(NA, n) if TRUE otherwise rep(length(unique(x)), n). Therefore apply will output a matrix. data$output <- apply(... tries assign a matrix (your result) into a column in your data.frame, data$output. This is the cause of your error.
Your code will run if you just assign your output to a variable instead
out <- apply(data, 1, function(x) {ifelse(x == 0, NA, length(unique(x)))})
If you are not expecting a class(matrix) as your output, but rather a vector, then there is something wrong with the logic of your function.
