Applying a threshold in a row-wise fashion, where cutoff value is relative to sd of row

Question

I have a matrix similar to this:

m <- matrix(rnorm(100), 100, 50)

and I'd like to change all the values row by row, so that any value above its row (standard deviation) *2 will become 1, otherwise 0 (basically a threshold).

I tried something like this:

cutoff <- function(x){
  x[x < 2*sd(x)] <- 0
  x[x > 2*sd(x)] <- 1
  return(x)
} 

mT <- apply(m, 1, cutoff)

but it's giving me something different. Any help would be very appreciated.

Ronak Shah · Accepted Answer · 2021-06-15T08:07:40.517

1

Your code is correct, you need to transpose the result as apply always returns a transposed result (See Why apply() returns a transposed xts matrix? ).

mT <- t(apply(m, 1, cutoff))

You can also reduce the cutoff function to -

cutoff <- function(x){
  as.integer(x > 2*sd(x))
}

x > 2*sd(x) returns a logical value (TRUE/FALSE), if we convert it to integer it changes TRUE to 1 and FALSE to 0.

edited Jun 15 '21 at 08:07

answered Jun 15 '21 at 07:39

Ronak Shah

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Thanks, it works now. I still don't understand why it needs transposing? I also don't understand how the as.integer function is working in this case – vankappa Jun 15 '21 at 08:00
I edited my answer to include some explanations for both the issues. – Ronak Shah Jun 15 '21 at 08:08

score 1 · Answer 2 · answered Jun 15 '21 at 07:47

1

Will this work:

t(apply(m, 1, function(x) +(x > 2*sd(x))))

answered Jun 15 '21 at 07:47

Karthik S

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score 1 · Answer 3 · answered Jun 15 '21 at 16:56

1

We can use dapply from collapse without transposing and is very efficient

library(collapse)
dapply(m, cutoff, MARGIN = 1)

answered Jun 15 '21 at 16:56

akrun

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Applying a threshold in a row-wise fashion, where cutoff value is relative to sd of row

3 Answers3