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Home/ Questions/Q 7565321
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Editorial Team
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Editorial Team
Asked: May 30, 20262026-05-30T14:07:33+00:00 2026-05-30T14:07:33+00:00

Consider the following matrix, nc <- 5000 nr <- 1024 m <- matrix(rnorm(nc*nr), ncol=nc)

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Consider the following matrix,

nc <- 5000
nr <- 1024
m <- matrix(rnorm(nc*nr), ncol=nc)

I wish to take the difference between the rowMeans of two groups of identical size taken at random in this matrix.

n <- 1000 # group size

system.time(replicate(100, {
   ind1 <- sample(seq.int(nc), n) 
   ind2 <- sample(seq.int(nc), n)
   rowMeans(m[, ind1]) - rowMeans(m[, ind2])
}))

It is quite slow, unfortunately I didn’t understand the output of Rprof (it seemed most of the time was spent on is.data.frame??)

Suggestions for something more efficient?

I have contemplated the following:

  • Rcpp: from my online readings I believe R’s rowMeans is quite efficient, so it’s not clear it would help at this step. I’d like to be convinced of where the bottleneck really is first, maybe my whole design is suboptimal. If most of the time is spent in making copies for each of the smaller matrices, would Rcpp perform better?

  • updating to R-devel, there seems to be a new .rowMeans function more efficient yet. Has anyone tried it?

Thanks.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-30T14:07:34+00:00Added an answer on May 30, 2026 at 2:07 pm

    Each rowSums() call on a subset of columns from m can be seen as the matrix multiplication between m and a vector of 0 or 1 indicating the selected columns. If you juxtapose all those vectors, you end up with a multiplication between two matrices (which is much more efficient):

    ind1 <- replicate(100, seq.int(nc) %in% sample(seq.int(nc), n)) 
    ind2 <- replicate(100, seq.int(nc) %in% sample(seq.int(nc), n))
    output <- m %*% (ind1 - ind2)
    
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