Use max on each element of a matrix

Question

> x <- array(-10:10, dim=c(4,5))
> x
     [,1] [,2] [,3] [,4] [,5]
[1,]  -10   -6   -2    2    6
[2,]   -9   -5   -1    3    7
[3,]   -8   -4    0    4    8
[4,]   -7   -3    1    5    9

How do I apply "max(x, 0)" to each element so that I get this matrix:

     [,1] [,2] [,3] [,4] [,5]
[1,]    0    0    0    2    6
[2,]    0    0    0    3    7
[3,]    0    0    0    4    8
[4,]    0    0    1    5    9

Solution 1

Use pmax:

pmax(x,0)
#     [,1] [,2] [,3] [,4] [,5]
#[1,]    0    0    0    2    6
#[2,]    0    0    0    3    7
#[3,]    0    0    0    4    8
#[4,]    0    0    1    5    9

Solution 2

You can use R's indexing function [ to do this directly:

x <- array(-10:10, dim=c(4,5))
x[x < 0] <- 0

This works because x < 0 creates a logical matrix output:

x < 0

     [,1] [,2]  [,3]  [,4]  [,5]
[1,] TRUE TRUE  TRUE FALSE FALSE
[2,] TRUE TRUE  TRUE FALSE FALSE
[3,] TRUE TRUE FALSE FALSE FALSE
[4,] TRUE TRUE FALSE FALSE FALSE

And the resulting matrix is:

     [,1] [,2] [,3] [,4] [,5]
[1,]    0    0    0    2    6
[2,]    0    0    0    3    7
[3,]    0    0    0    4    8
[4,]    0    0    1    5    9

The timing between the two methods is surprisingly similar. Here's a larger example illustrating the comparable timings:

xbigC <- xbigE <- matrix(sample(-100:100, 1e8, TRUE), ncol = 1e4)
system.time(xbigC[xbigC < 0] <- 0)
 #--- 
  user  system elapsed 
   4.56    0.37    4.93 
system.time(xbigE  <- pmax(xbigE,0))
#---
   user  system elapsed 
   4.10    0.51    4.62 
all.equal(xbigC, xbigE)
#---
[1] TRUE

Solution 3

It appears that the order of the arguments to pmax() affects the class of what is returned when the input is a matrix:

pmax(0,x)
[1] 0 0 0 0 0 0 0 0 0 0 0 1 2 3 4 5 6 7 8 9

pmax(x,0)
     [,1] [,2] [,3] [,4] [,5]
[1,]    0    0    0    2    6
[2,]    0    0    0    3    7
[3,]    0    0    0    4    8
[4,]    0    0    1    5    9

Comments

[Chase]

@ManInMoon - yep, no worries. The direct indexing approach will be more flexible in that you can create more complicated conditions with other logical operators. For this task, it seems pmax() is ever so slightly more efficient.

[Tim Kuipers]

This did not work for me. I did > x=read.csv(...) > x2 = sapply(data,atanh) > pmax(x2,-20) but this produced a vector of -20's instead of a matrix of the original dimensions. (Note that > class(x2) [1] "matrix" eventhough > class(x) [1] "data.frame") @Chase his solution did work.

[eddi]

@TimKuipers pmax preserves the dim of the object - you're probably leaving out a step or your x2 is not rectangular

[Tim Kuipers]

Ah no. The mistake I made was that since my matrix was really big, I tried looking at just a part of it by doing x2[1:10] instead of x2[1:10,]. I'm sorry!

[eddi]

@TimKuipers np, glad you figured it out