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Different hard threshold for each column

Tags:

r

matrix

I would like to hard threshold my matrix such that all values below a certain number are set to zero. However, I would like that threshold to vary by the column (i.e. each column has its own threshold). How can I do this in R?

Here is the simple set up:

set.seed(1)
A <- matrix(runif(n = 12),nrow = 4)
#    [,1]       [,2]      [,3]
#[1,] 0.2655087 0.2016819 0.62911404
#[2,] 0.3721239 0.8983897 0.06178627
#[3,] 0.5728534 0.9446753 0.20597457
#[4,] 0.9082078 0.6607978 0.17655675



threshholds <- c(0.3,1,0.5)

#wanted result: 

#    [,1]       [,2]      [,3]
#[1,] 0         0         0.62911404
#[2,] 0.3721239 0         0        
#[3,] 0.5728534 0         0        
#[4,] 0.9082078 0         0        

I need to apply it to large matrices, so efficiency is relevant.


Edit: Having received several excellent suggestions, I compared their speed for future reference:
set.seed(1)
A <- matrix(runif(n = 1E4*2E3),nrow = 2E3)

threshholds <- runif(n=1E4)

> system.time(A * (A > threshholds[col(A)]))# akrun
   user  system elapsed 
  0.394   0.124   0.519 
> system.time(replace(A, A <= threshholds[col(A)], 0)) # akrun
   user  system elapsed 
  0.465   0.138   0.604 
> system.time(pmin(A, A > threshholds[col(A)])) #akrun
   user  system elapsed 
  0.678   0.290   1.024 
> system.time(A[t(apply(A, 1, `<`, threshholds))] <- 0) #Andrew Gustar
   user  system elapsed 
  0.875   0.306   1.200 
> system.time(At <- apply(A, 1, applythresh)) + system.time(t(At)) #Chris Litter
   user  system elapsed 
  0.891   0.372   1.286 
> system.time(sweep(A, 2, threshholds, function(a,b) {ifelse(a<b,0,a)})) #MrFlick
   user  system elapsed 
  1.752   0.598   2.354 
like image 512
The_Anomaly Avatar asked Jun 15 '18 19:06

The_Anomaly


Video Answer


1 Answers

Here is a vectorized option

replace(A, A <= threshholds[col(A)], 0)

Or with some arithmetic

A * (A > threshholds[col(A)])
#       [,1] [,2]     [,3]
#[1,] 0.0000000    0 0.629114
#[2,] 0.3721239    0 0.000000
#[3,] 0.5728534    0 0.000000
#[4,] 0.9082078    0 0.000000

Or with pmin

pmin(A, A > threshholds[col(A)])
#         [,1] [,2]     [,3]
#[1,] 0.0000000    0 0.629114
#[2,] 0.3721239    0 0.000000
#[3,] 0.5728534    0 0.000000
#[4,] 0.9082078    0 0.000000
like image 108
akrun Avatar answered Sep 28 '22 00:09

akrun