Using "apply" to apply a function to a matrix where parameters are column-specific

Question

I'm trying to avoid using loops by using apply to apply a user-defined function to a matrix. The problem I have is that there are additional parameters that my function uses and they differ for each column of the matrix. Below is a toy example.

Say I have the following function:

foo <- function(x, a, b, c) return( (a*x + b)^c )

and I want to apply it to a matrix bar using different values of a, b, and c for each column.

bar <- matrix(1:15, ncol = 3)
a <- 4:6
b <- 3:1
c <- 1:3

In this case, for the first column of bar, then a=4, b=3, and c=1. I tried this,

apply(bar, 2, foo, a=a, b=b, c=c)

but this is clearly incorrect, as each column uses all parameters sequentially before wrapping back to the first parameter again. Any suggestions?

akrun · Accepted Answer · 2015-05-15T17:30:49.077

12

We could split the 'bar' by 'column' (col(bar)) and with mapply we can apply 'foo' for the corresponding 'a', 'b', 'c' values to each column of 'bar'

mapply(foo, split(bar, col(bar)), a, b, c)

Or without using apply

ind <- col(bar)
(a[ind]*bar +b[ind])^c[ind]

edited May 15 '15 at 17:30

answered May 15 '15 at 17:17

akrun

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Instead of `split` you can use also `as.data.frame`. – nicola May 15 '15 at 17:32
@nicola Yes, I thought to keep it in the `matrix` – akrun May 15 '15 at 17:33
2

I'd already upvoted the first one and now wish there were the ability to add votes for exceptionally elegant edits. – IRTFM May 15 '15 at 17:36

score 9 · Answer 2 · answered May 15 '15 at 17:38

9

I don't see why you bother with a function at all:

> a <- matrix(4:6,nrow = 5,ncol = 3,byrow = TRUE)
> b <- matrix(3:1,nrow = 5,ncol = 3,byrow = TRUE)
> c <- matrix(1:3,nrow = 5,ncol = 3,byrow = TRUE)
> (a*bar + b)^c
     [,1] [,2]   [,3]
[1,]    7 1024 300763
[2,]   11 1369 389017
[3,]   15 1764 493039
[4,]   19 2209 614125
[5,]   23 2704 753571

answered May 15 '15 at 17:38

joran

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@akrun Ah, I missed the end of your answer. – joran May 15 '15 at 17:40
It's okay. I wouldn't have guessed it anyway – akrun May 15 '15 at 17:41
@akrun I actually think your use of `col` is much better. – joran May 15 '15 at 17:42
I think `rep` might be a bit more faster compared to recycling and `col` – akrun May 15 '15 at 17:43
@joran The code I give above is just a toy example. The real problem uses a much more complex function that was defined by someone else in an R package – specifically `tran.1D` in the `ReacTran` package. – Dan May 15 '15 at 18:17

score 3 · Answer 3 · answered May 15 '15 at 17:42

3

I guess that you can just transpose your matrix and use vectorization:

t(foo(t(bar),a,b,c))

This should work for every vectorized foo.

answered May 15 '15 at 17:42

nicola

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score 2 · Answer 4 · answered May 15 '15 at 17:25

You could put your parameters into the vector:

newbar <- rbind(a,b,c,bar)
newfoo <- function(z){x <- z[-(1:3)]; (z[1]*x+z[2])^z[3]}
apply(newbar,2,newfoo)

which gives

[,1] [,2]   [,3]
    7 1024 300763
   11 1369 389017
   15 1764 493039
   19 2209 614125
   23 2704 753571

score 2 · Answer 5 · answered May 15 '15 at 17:34

2

You can use sweep; usually this is for subtracting the mean from each column, but you can pass in an index instead to get parallel indexing across a,b and c:

> sweep(bar, 2, seq_along(a), function(x,i) foo(x, a[i], b[i], c[i]), FALSE)
     [,1] [,2]   [,3]
[1,]    7 1024 300763
[2,]   11 1369 389017
[3,]   15 1764 493039
[4,]   19 2209 614125
[5,]   23 2704 753571

answered May 15 '15 at 17:34

Neal Fultz

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See also http://stackoverflow.com/questions/3444889/how-to-use-the-r-function-sweep – Neal Fultz May 15 '15 at 17:36

Using "apply" to apply a function to a matrix where parameters are column-specific

5 Answers5