djmuseR wrote:
>
> Hi:
>
> To take this idea a step further,
>
> colSums(apply(unique(A), 1, function(x) apply(A, 1, function(y)
> identical(x, y))))
> [1] 10 5 5
>
> However, Denis' solution is a bit faster because fewer evaluations are
> required:
>
>> system.time(replicate(1000,
> + colSums(apply(unique(A), 1, function(x) apply(A, 1, function(y)
> identical(x, y))))
> + ))
> user system elapsed
> 1.42 0.00 1.42
>> system.time(replicate(1000, {B <- apply(A,1, FUN = function(x)paste(x,
>> collapse = ''))
> + table(B) }))
> user system elapsed
> 0.81 0.00 0.81
>
> system.time(replicate(1000,{sum(apply(A,1, FUN =
> function(x)identical(x,x1)))}))
user system elapsed
0.301 0.001 0.305
>
> system.time(replicate(1000,{sum(apply(A,1, FUN =
> function(x)all(zapsmall(x-x1)==0)))}))
user system elapsed
0.672 0.002 0.675
>
> system.time(replicate(1000, {B <- apply(A,1, FUN = function(x)paste(x,
> collapse = '')) ; table(B) }))
user system elapsed
0.835 0.002 0.837
Berend
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