R: combining output from apply for each value of an array dimension

Question

I hope to obtain the percentage of rows with a maximum value of 1 in a 4-dimensional array. I can obtain that information using an apply function for individual values of the 4th dimension, then use cbind to combine all of the output. However, the number of values in the 4th dimension may vary. Is there an efficient way to obtain the output in the desired format without using cbind?

Below is functional code that returns the desired information as separate vectors for each level of the 4th dimension of an array.

set.seed(9345)

A <- 8
B <- 2
C <- 5
D <- 2  # the value of D might change

y2 <- array(rbinom(A*B*C*D, 1, 0.4), dim = c(A, B, C, D))

pA <- colMeans(apply(y2[,,,1], c(1,3), max))
pB <- colMeans(apply(y2[,,,2], c(1,3), max))

pp <- cbind(pA, pB)
pp

#        pA    pB
# [1,] 0.750 0.875
# [2,] 0.625 0.250
# [3,] 0.375 1.000
# [4,] 0.375 0.500
# [5,] 0.625 0.750

Is there an easy way to obtain the equivalent of pp without using cbind on the output pA and pB (maybe a one-liner that incorporates the apply and colMeans functions)? If the number of values in the 4th dimension (D) change then the number of vectors that must be combined with cbind would vary.

Thank you for any suggestions.

The code you've posted doesn't work without editing. (You set up an object y, but then operate on y2.) If you do y2 <- y, then the resultant matrix isn't as shown. (It instead matches the one in my answer below). — Josh O'Brien
– Josh O'Brien, Commented Apr 24, 2012 at 23:44

Josh O'Brien · Accepted Answer · 2012-04-25 00:43:48Z

4

This should be equivalent, and simpler to boot:

colMeans(apply(y2[,,,], c(1,3,4), max))
#       [,1]  [,2]
# [1,] 0.750 0.875
# [2,] 0.625 0.250
# [3,] 0.375 1.000
# [4,] 0.375 0.500
# [5,] 0.625 0.750

edited Apr 25, 2012 at 0:43

answered Apr 24, 2012 at 23:40

Josh O'Brien

163k29 gold badges380 silver badges465 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

Mark Miller Over a year ago

Thanks for the answer. I am thinking if I use c(1,3,4) I do not need the t(): colMeans(apply(y2[,,,], c(1,3,4), max)).

Josh O'Brien Over a year ago

@MarkMiller -- Thanks for catching that (fixed now). (It was left over from the experiments I'd run when I was puzzling over why c(1,3,4) didn't match your posted results!)

Collectives™ on Stack Overflow

R: combining output from apply for each value of an array dimension

1 Answer 1

2 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

2 Comments

Your Answer

Sign up or log in

Post as a guest

Related