[R] Computing sums of the columns of an array

[Uwe Ligges]

Duncan Murdoch wrote:

> 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))
> [1] 0.08 0.00 0.08   NA   NA
>  >   system.time(rep(1, nrow(A)) %*% A)
> [1] 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))
> [1] 0.09 0.00 0.09   NA   NA

On my machine:
[1] 0.12 0.00 0.12   NA   NA

>  >   system.time(rep(1, nrow(A)) %*% A)
> [1] 0.11 0.00 0.12   NA   NA

On my machine:
[1] 0.32 0.00 0.32   NA   NA


Hence still a bigger factor both with R-2.1.1 release (standard BLAS; 
gcc-3.4.2, WinNT4.0, Athlon XP with real freq. of 1667MHz, 1Gb).
And still a bigger factor (0.09 vs. 0.21) on a Xeon 3.06Ghz with 2Gb.

Are you using Goto's BLAS (for me, your performance is still not 
achievable with ATLAS)?

Best,
Uwe


> Still, colSums is the winner in both cases.
> 
> Duncan Murdoch