Hi Jim,
thanks that works like a charm.
cheers
Peter
> On 5. Jul 2017, at 12:01, Jim Lemon <drjimle...@gmail.com> wrote:
>
> Hi Peter,
>
> apply(t(apply(mm,1,rep,each=3)),2,rep,each=3)
>
> Jim
>
> On Wed, Jul 5, 2017 at 5:20 PM, Anthoni, Peter (IMK)
> <peter.anth...@kit.edu> wrote:
>> Hi all,
>> (if me email goes out as html, than my email client don't do as told, and I
>> apologies already.)
>>
>> We need to downscale climate data and therefore first need to expand the
>> climate from 0.5deg to the higher resolution 10min, before we can add high
>> resolution deviations. We basically need to have the original data at each
>> gridcell replicated into 3x3 gridcells.
>> A simple for loop can do this, but I could need a faster procedure. Anybody
>> know a faster way? Is there package than can do what we need already?
>> I tried matrix with rep, but I am missing some magic there, since it doesn't
>> do what we need.
>> replicate might be promising, but then still need to rearrange the output
>> into the column and row format we need.
>>
>> A simple example:
>> mm=matrix(1:15,nrow=3,byrow = T)
>> xmm=matrix(NA,nrow=nrow(mm)*3,ncol=ncol(mm)*3)
>> for(icol in 1:ncol(mm)) {
>> for(irow in 1:nrow(mm)) {
>> xicol=(icol-1)*3 +c(1:3)
>> xirow=(irow-1)*3 +c(1:3)
>> xmm[xirow,xicol]=mm[irow,icol]
>> }
>> }
>> mm
>>>> mm
>>> [,1] [,2] [,3] [,4] [,5]
>>> [1,] 1 2 3 4 5
>>> [2,] 6 7 8 9 10
>>> [3,] 11 12 13 14 15
>>>
>> xmm
>>>> xmm
>>> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13]
>>> [,14] [,15]
>>> [1,] 1 1 1 2 2 2 3 3 3 4 4 4 5
>>> 5 5
>>> [2,] 1 1 1 2 2 2 3 3 3 4 4 4 5
>>> 5 5
>>> [3,] 1 1 1 2 2 2 3 3 3 4 4 4 5
>>> 5 5
>>> [4,] 6 6 6 7 7 7 8 8 8 9 9 9 10
>>> 10 10
>>> [5,] 6 6 6 7 7 7 8 8 8 9 9 9 10
>>> 10 10
>>> [6,] 6 6 6 7 7 7 8 8 8 9 9 9 10
>>> 10 10
>>> [7,] 11 11 11 12 12 12 13 13 13 14 14 14 15
>>> 15 15
>>> [8,] 11 11 11 12 12 12 13 13 13 14 14 14 15
>>> 15 15
>>> [9,] 11 11 11 12 12 12 13 13 13 14 14 14 15
>>> 15 15
>>
>> I tried various rep with matrix, but don't get the right result.
>> xmm2=matrix(rep(rep(mm,each=3),times=3),nrow=nrow(mm)*3,ncol=ncol(mm)*3,byrow
>> = F)
>>> identical(xmm,xmm2)
>> [1] FALSE
>>
>> rr=replicate(3,rep(t(mm),each=3))
>> rr
>>>> rr
>>> [,1] [,2] [,3]
>>> [1,] 1 1 1
>>> [2,] 1 1 1
>>> [3,] 1 1 1
>>> [4,] 2 2 2
>>> [5,] 2 2 2
>>> [6,] 2 2 2
>>> [7,] 3 3 3
>>> ...
>> identical(xmm,matrix(rr,ncol=15,nrow=9,byrow=T))
>>>> identical(xmm,matrix(rr,ncol=15,nrow=9,byrow=T))
>>> [1] FALSE
>>
>> Many thanks for any advice.
>>
>> cheers
>> Peter
>>
>> ______________________________________________
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