I like David's answer and it can be made much faster.
I show three refinements, each faster than the preceding one.
Rich
> system.time(for (i in 1:1000)
+ mat[seq(1, nrow(mat), by=2), ]+mat[seq(2, nrow(mat), by=2), ]
+ )
user system elapsed
0.18 0.00 0.19
>
> system.time(for (i in 1:1000)
+ mat[seqn <- seq(1, nrow(mat), by=2), ]+mat[seqn+1, ]
+ )
user system elapsed
0.08 0.00 0.08
>
> system.time(for (i in 1:1000)
+ mat[seqn <- seq(1, length=nrow(mat)/2, by=2), ]+mat[seqn+1, ]
+ )
user system elapsed
0.05 0.00 0.05
>
> system.time(for (i in 1:1000)
+ {mat2 <- mat; dim(mat2) <- c(2,3,6); mat2[1,,]+mat2[2,,]}
+ )
user system elapsed
0.01 0.00 0.02
>
On Fri, Apr 22, 2011 at 12:28 PM, David Winsemius <dwinsem...@comcast.net>wrote:
>
> On Apr 22, 2011, at 12:13 PM, Christine SINOQUET wrote:
>
> Hello,
>>
>> mat1 only consists of 0s and 1s:
>> 0 0 1 0 0 0
>> 1 1 0 1 1 0
>> 1 1 1 0 1 0
>> 0 1 1 0 0 1
>> 1 0 0 1 0 0
>> 0 1 0 1 0 1
>>
>> N = 3
>> M = 6
>>
>> I would like to "compress" mat1 every two rows, applying summation over
>> the two rows (per column), at each step, to yield:
>>
>> mat2
>> 1 1 1 1 1 0
>> 1 2 2 0 1 1
>> 1 1 0 2 0 1
>>
>
> > mat[seq(1, nrow(mat), by=2), ]+mat[seq(2, nrow(mat), by=2), ]
> [,1] [,2] [,3] [,4] [,5] [,6]
> [1,] 1 1 1 1 1 0
> [2,] 1 2 2 0 1 1
> [3,] 1 1 0 2 0 1
>
>
>
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