[R] filling the matrix row by row in the order from lower to larger elements

[Bert Gunter]

Preoccupation with speed of execution is typically (certainly not
always) misplaced. Provided you have used sensible basic
vectorization, loops in whatever form work adequately. First get
working code. Then, if necessary, parallelization, byte compilation,
or complex vectorization strategies can be employed. And, of course,
there's always C code for those who know it.

-- Bert

On Sun, Apr 8, 2012 at 7:31 AM, Dimitri Liakhovitski
<dimitri.liakhovitski at gmail.com> wrote:
> Thank you very much, Rui.
> This definitely produces the result needed.
> Again, I have not checked the speed yet.
>
> input<-as.matrix(data.frame(a=c(5,1,3,7),b=c(2,6,4,8)))
> f <- function(x){
>       nr <- nrow(x)
>       result <- matrix(0, nrow=nr, ncol=ncol(x))
>       colnames(result) <- colnames(x)
>
>       inp.ord <- order(x)[1:nr] - 1 # Keep only one per row, must be zero based
>       inx <- cbind(1:nr, inp.ord %/% nr + 1) # Index matrix
>       result[inx] <- 100
>       result
> }
> f(input)
>
> Dimitri
>
> On Fri, Apr 6, 2012 at 7:02 PM, Rui Barradas <rui1174 at sapo.pt> wrote:
>> Hello,
>>
>> Oops!
>>
>> What happened to the function 'f'?
>> Forgot to copy and pasted only the rest, now complete.
>>
>>
>> f <- function(x){
>>        nr <- nrow(x)
>>        result <- matrix(0, nrow=nr, ncol=ncol(x))
>>        colnames(result) <- colnames(x)
>>
>>        inp.ord <- order(x)[1:nr] - 1 # Keep only one per row, must be zero based
>>        inx <- cbind(1:nr, inp.ord %/% nr + 1) # Index matrix
>>        result[inx] <- 100
>>        result
>> }
>>
>> # Original example
>> input <- as.matrix(data.frame(a=c(5,1,3,7), b=c(2,6,4,8)))
>> (input)
>> desired.result <- as.matrix(data.frame(a=c(100,0,100,0), b=c(0,100,0,100)))
>> (desired.result)
>> all.equal(f(input), desired.result)
>>
>> # Two other examples
>> set.seed(123)
>> (x <- matrix(sample(10, 10), ncol=2))
>> f(x)
>>
>> (y <- matrix(sample(40, 40), ncol=5))
>> f(y)
>>
>> Note that there's no loops (or apply, which is also a loop.)
>>
>> Rui Barradas
>>
>>
>> --
>> View this message in context: http://r.789695.n4.nabble.com/filling-the-matrix-row-by-row-in-the-order-from-lower-to-larger-elements-tp4538171p4538486.html
>> Sent from the R help mailing list archive at Nabble.com.
>>
>> ______________________________________________
>> R-help at r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>
>
>
> --
> Dimitri Liakhovitski
> marketfusionanalytics.com
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.



-- 

Bert Gunter
Genentech Nonclinical Biostatistics

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