# Re: [R] Computing sums of the columns of an array

From: Duncan Murdoch <murdoch_at_stats.uwo.ca>
Date: Sat 06 Aug 2005 - 02:55:06 EST

On 8/5/2005 12:43 PM, Uwe Ligges wrote:

```> Duncan Murdoch wrote:
>
>> On 8/5/2005 12:16 PM, Martin C. Martin wrote:
>>
```

>>>Hi,
>>>
>>>I have a 5x731 array A, and I want to compute the sums of the columns.
>>>Currently I do:
>>>
>>>apply(A, 2, sum)
>>>
>>>But it turns out, this is slow: 70% of my CPU time is spent here, even
>>>though there are many complicated steps in my computation.
>>>
>>>Is there a faster way?
```>>
>>
>> You'd probably do better with matrix multiplication:
>>
>> rep(1, nrow(A)) %*% A
>
>
> No, better use colSums(), which has been optimized for this purpose:
>
>   A <- matrix(seq(1, 10000000), ncol=10000)
>   system.time(colSums(A))
>   # ~ 0.1 sec.
>   system.time(rep(1, nrow(A)) %*% A)
>   # ~ 0.5 sec.

```

I didn't claim my solution was the best, only better. :-)

One point of interest: I think your example exaggerates the difference by using a matrix of integers. On my machine I get a ratio something like yours with the same example

> A <- matrix(seq(1, 10000000), ncol=10000)  > system.time(colSums(A))
 0.08 0.00 0.08 NA NA
> system.time(rep(1, nrow(A)) %*% A)
 0.25 0.01 0.23 NA NA

but if I make A floating point, there's much less difference:

> A <- matrix(as.numeric(seq(1, 10000000)), ncol=10000)  > system.time(colSums(A))
 0.09 0.00 0.09 NA NA
> system.time(rep(1, nrow(A)) %*% A)
 0.11 0.00 0.12 NA NA

Still, colSums is the winner in both cases.

Duncan Murdoch

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