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
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.anthoni at 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 > > ______________________________________________ > R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see > 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.
Hi Jim, thanks that works like a charm. cheers Peter> On 5. Jul 2017, at 12:01, Jim Lemon <drjimlemon at 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.anthoni at 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 >> >> ______________________________________________ >> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see >> 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.
You probably ought to be using the raster package. See the CRAN Spatial Task View. -- Sent from my phone. Please excuse my brevity. On July 5, 2017 12:20:28 AM PDT, "Anthoni, Peter (IMK)" <peter.anthoni at 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 > >______________________________________________ >R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see >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.
Hi Jeff, thanks, the raster package disaggregate will do the trick as well. library(raster) rmm <- raster(ncols=5, nrows=3) rmm[] <- matrix(1:15,nrow=3,byrow = T) xrmm <- disaggregate(rmm, fact=c(3, 3))> > as.matrix(rmm) > [,1] [,2] [,3] [,4] [,5] > [1,] 1 2 3 4 5 > [2,] 6 7 8 9 10 > [3,] 11 12 13 14 15 > > as.matrix(xrmm) > [,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 15the disaggregate as a bit faster than the tapply. mmb=matrix(1:259200,nrow=720,ncol=360) rmmb <- raster(ncols=360, nrows=720) rmmb[] <- mmb[] system.time(for(i in 1:10) {xmm=matrix(NA,nrow=nrow(mmb)*3,ncol=ncol(mmb)*3) for(icol in 1:ncol(mmb)) { for(irow in 1:nrow(mmb)) { xicol=(icol-1)*3 +c(1:3) xirow=(irow-1)*3 +c(1:3) xmm[xirow,xicol]=mmb[irow,icol] } } }) system.time(for(i in 1:10) {apply(t(apply(mmb,1,rep,each=3)),2,rep,each=3)}) #ca. 10x faster system.time(for(i in 1:10) {xrmmb <- disaggregate(rmmb, fact=c(3, 3))})> > system.time(for(i in 1:10) {xmm=matrix(NA,nrow=nrow(mmb)*3,ncol=ncol(mmb)*3) > + for(icol in 1:ncol(mmb)) { > + for(irow in 1:nrow(mmb)) { > + xicol=(icol-1)*3 +c(1:3) > + xirow=(irow-1)*3 +c(1:3) > + xmm[xirow,xicol]=mmb[irow,icol] > + } > + } > + }) > user system elapsed > 8.297 0.048 8.364 > > system.time(for(i in 1:10) {apply(t(apply(mmb,1,rep,each=3)),2,rep,each=3)}) #ca. 10x faster > user system elapsed > 0.785 0.093 0.881 > > system.time(for(i in 1:10) {xrmmb <- disaggregate(rmmb, fact=c(3, 3))}) > user system elapsed > 0.583 0.147 0.731cheers Peter> On 5. Jul 2017, at 16:57, Jeff Newmiller <jdnewmil at dcn.davis.ca.us> wrote: > > You probably ought to be using the raster package. See the CRAN Spatial Task View. > -- > Sent from my phone. Please excuse my brevity. > > On July 5, 2017 12:20:28 AM PDT, "Anthoni, Peter (IMK)" <peter.anthoni at 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 >> >> ______________________________________________ >> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see >> 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.