Row-wise variance of a matrix in R

Question

I'd like to compute the variance for each row in a matrix. For the following matrix A

     [,1] [,2] [,3]
[1,]    1    5    9
[2,]    5    6   10
[3,]   50    7   11
[4,]    4    8   12

I would like to get

[1]  16.0000   7.0000 564.3333  16.0000

I know I can achieve this with apply(A,1,var) , but is there a faster or better way? From octave, I can do this with var(A,0,2) , but I don't get how the Y argument of the var() function in R is to be used.

Edit: The actual dataset of a typical chunk has around 100 rows and 500 columns. The total amount of data is around 50GB though.

Answer 1

You could potentially vectorize var over rows (or columns) using rowSums and rowMeans

RowVar <- function(x, ...) {
  rowSums((x - rowMeans(x, ...))^2, ...)/(dim(x)[2] - 1)
}

RowVar(A)
#[1]  16.0000   7.0000 564.3333  16.0000

Using @Richards data, yields in

microbenchmark(apply(m, 1, var), RowVar(m))

## Unit: milliseconds
## expr        min         lq     median         uq        max neval
## apply(m, 1, var) 343.369091 400.924652 424.991017 478.097573 746.483601   100
##        RowVar(m)   1.766668   1.916543   2.010471   2.412872   4.834471   100

You can also create a more general function that will receive a syntax similar to apply but will remain vectorized (the column wise variance will be slower as the matrix needs to be transposed first)

MatVar <- function(x, dim = 1, ...) {
  if(dim == 1){
     rowSums((x - rowMeans(x, ...))^2, ...)/(dim(x)[2] - 1)
  } else if (dim == 2) {
     rowSums((t(x) - colMeans(x, ...))^2, ...)/(dim(x)[1] - 1)
  } else stop("Please enter valid dimension")
}


MatVar(A, 1)
## [1]  16.0000   7.0000 564.3333  16.0000

MatVar(A, 2)
        V1         V2         V3 
## 547.333333   1.666667   1.666667 

Answer 2

This is one of the main reasons why apply() is useful. It is meant to operate on the margins of an array or matrix.

set.seed(100)
m <- matrix(sample(1e5L), 1e4L)
library(microbenchmark)
microbenchmark(apply(m, 1, var))
# Unit: milliseconds
#              expr      min       lq   median       uq      max neval
#  apply(m, 1, var) 270.3746 283.9009 292.2933 298.1297 343.9531   100 

Is 300 milliseconds too long to make 10,000 calculations?