Hi there,
Thanks again for your reply. I know for-loop is always a solution to my
problem and I had already coded using for-loop. But the number of levels for
each dimension is large enough in actual problem and hence it was
time-consuming.
So, I was just wondering if there are any other alternative way-outs to
solving my problem. That's why I tried with apply functions (sapply)assuming
that this might work out faster even fractionally as compared to for-loop.
Cheers,
Sauvik
On Mon, Jul 27, 2009 at 12:28 AM, Poersching <poerschin...@web.de> wrote:
> Sauvik De schrieb:
>
> Hi:
> Lots of thanks for your valuable time!
>
> But I am not sure how you would like to use the function in this situation.
>
> As I had mentioned that the first element of my output array should be
> like:
>
>
> cor(DataArray_1[dimnames(Correl)[[1]][1],dimnames(Correl)[[2]][1],dimnames(Correl)[[4]][1],],DataArray_2[dimnames(Correl)[[1]][1],dimnames(Correl)[[3]][1],dimnames(Correl)[[4]][1],],use="pairwise.complete.obs")
>
> in my below code.
>
> and
>
> the output array of correlation I wish to get using "sapply" as follows:
>
> Correl = sapply(Correl,function(d) cor(DataArray_1[...],DataArray_2[...],
> use="pairwise.complete.obs"))
>
> So it would be of great help if you could kindly specify how to utilise
> your function "findIndex" in ...
>
> Apologies for all this!
>
> Thanks & Regards,
> Sauvik
>
> Hey,
> sorry, I haven't understood your problem last time, but now this solution
> should solve your problem, so I hope. :-)
> It's only a for to loop, but an apply function may work too. I will think
> about this, but for now... ;-)
>
> la<-length(a)
> lb<-length(b)
> lc<-length(c)
> ld<-length(d)
> for (ia in 1:la) {
> for (ib in 1:lb) {
> for (ic in 1:lc) {
> for (id in 1:ld) {
> Correl[ia,ib,ic,id]<-cor(
> DataArray_1[dimnames(Correl)[[1]][ia],
> dimnames(Correl)[[2]][ib],
> dimnames(Correl)[[4]][id],]
> ,
> DataArray_2[dimnames(Correl)[[1]][ia],
> dimnames(Correl)[[3]][ic],
> dimnames(Correl)[[4]][id],]
> ,
> use="pairwise.complete.obs")
> }
> }
> }
> }
> ## with function findIndex you can find the dimensions with
> ## i.e. cor values greater 0.5 or smaller -0.5, like:
> findIndex(Correl,Correl[Correl>0.5])
> findIndex(Correl,Correl[Correl<(-0.5)])
>
> I have changed the code of the function findIndex in line which contents:
> el[j]<-which(is.element(data,element[j]))
>
> Rigards,
> Christian
>
>
> On Sun, Jul 26, 2009 at 3:54 PM, Poersching<poerschin...@web.de> wrote:
> > Sauvik De schrieb:
> >
> > Hi Gabor:
> > Many thanks for your prompt reply!
> > The code is fine. But I need it in more general form as I had mentioned
> that
> > I need to input any 0 to find its dimension-names.
> >
> > Actually, I was using "sapply" to calculate correlation and this idea was
> > required in the middle of correlation calculation.
> > I am providing the way I tried my calculation.
> >
> > a= c("A1","A2","A3","A4","A5")
> > b= c("B1","B2","B3")
> > c= c("C1","C2","C3","C4")
> > d= c("D1","D2")
> > e= c("E1","E2","E3","E4","E5","E6","E7","E8")
> >
> > DataArray_1 = array(c(rnorm(240)),dim=c(length(a),length(b),
> > length(d),length(e)),dimnames=list(a,b,d,e))
> > DataArray_2 = array(c(rnorm(320)), dim=c(length(a),length(c),
> > length(d),length(e)),dimnames=list(a,c,d,e))
> >
> > #Defining an empty array which will contain the correlation values
> (output
> > array)
> > Correl = array(NA, dim=c(length(a),length(b),
> > length(c),length(d)),dimnames=list(a,b,c,d))
> >
> > #Calculating Correlation between attributes b & c over values of e
> > Correl = sapply(Correl,function(d) cor(DataArray_1[...],DataArray_2[...],
> > use="pairwise.complete.obs"))
> >
> > This is where I get stuck.
> > In the above, d is acting as an element in the "Correl" array. Hence I
> need
> > to get the dimension-names for d.
> >
> > #The first element of Correl will be:
> >
> cor(DataArray_1[dimnames(Correl)[[1]][1],dimnames(Correl)[[2]][1],dimnames(Correl)[[4]][1],],DataArray_2[dimnames(Correl)[[1]][1],dimnames(Correl)[[3]][1],dimnames(Correl)[[4]][1],],use="pairwise.complete.obs")
> >
> > So my problem boils down to extracting the dim-names in terms of
> element(d)
> > and not in terms of Correl (that I have mentioned as "..." in the above
> > code)
> >
> > My sincere thanks for your valuable time & suggestions.
> >
> > Many Thanks & Kind Regards,
> > Sauvik
> >
> >
> > On Sun, Jul 26, 2009 at 5:26 AM, Gabor Grothendieck <
> ggrothendi...@gmail.com
> >
> >
> > wrote:
> >
> >
> >
> >
> > Try this:
> >
> >
> >
> > ix <- c(1, 3, 4, 2)
> > mapply("[", dimnames(mydatastructure), ix)
> >
> >
> > [1] "S1" "T3" "U4" "V2"
> >
> >
> > On Sat, Jul 25, 2009 at 5:12 PM, Sauvik De<sauvik.s...@gmail.com> wrote:
> >
> >
> > Hi:
> > How can I extract the dimension-names of a pre-defined element in a
> > multidimensional array in R ?
> >
> > A toy example is provided below:
> > I have a 4-dimensional array with each dimension having certain length.
> >
> >
> > In
> >
> >
> > the below example, "mydatastructure" explains the structure of my data.
> >
> > mydatastructure = array(0,
> >
> >
> > dim=c(length(b),length(z),length(x),length(d)),
> >
> >
> > dimnames=list(b,z,x,d))
> >
> > where,
> > b=c("S1","S2","S3","S4","S5")
> > z=c("T1","T2", "T3")
> > x=c("U1","U2","U3","U4")
> > d=c("V1","V2")
> >
> > Clearly, "mydatastructure" contains many 0's.
> > Now how can I get the dimension-names of any particular 0 ?
> > That is, my input should be a particular 0 in the array "mydatastructure"
> > (Suppose this 0 corresponds to S1,T3,U4 & V2 in the array). Then my
> >
> >
> > output
> >
> >
> > should be S1,T3,U4 & V2.
> >
> > The function "dimnames" didn't help me with the solution.
> > Any idea will greatly be appreciated.
> >
> > Thanks for your time!
> >
> > Kind Regards,
> > Sauvik
> >
> > [[alternative HTML version deleted]]
> >
> > ______________________________________________
> > R-help@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.
> >
> >
> >
> > [[alternative HTML version deleted]]
> >
> > ______________________________________________
> > R-help@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.
> >
> >
> >
> > Hey,
> >
> > I have spend some time to write a function, which should fulfill your
> needs.
> > so i hope ;-)
> >
> > findIndex<-function(data,element) {
> > ld<-length(data)
> > el<-which(is.element(data,element))
> > lel<-length(el)
> > ndim<-length(dim(data))
> > ind<-array(,dim=c(lel,ndim),dimnames=list(el,1:ndim))
> > precomma<-""
> > tempdata<-data
> > tempel<-el
> > for (j in 1:lel) {
> > data<-tempdata
> > el<-tempel
> > ld<-length(data)
> > for (i in ndim:1) {
> > ratio<-el[j]/(ld/dim(data)[i])
> > if (ratio-trunc(ratio)>0) {
> > ind[j,i]<-trunc(ratio)+1
> > } else {
> > ind[j,i]<-trunc(ratio)
> > }
> > if (length(dim(data))>1) {
> > k<-1
> > while (k>=1 & k<=(i-1)) {
> > precomma<-paste(precomma,",",sep="")
> > k<-k+1
> > }
> >
> >
> data<-as.array(eval(parse(text=paste("data[",precomma,ind[j,i],"]",sep=""))))
> > precomma<-""
> > ld<-length(data)
> > el[j]<-which(is.element(data,element[j]))
> > }
> > }
> > }
> > return(ind)
> > }
> >
> > Regards,
> > Christian Porsche
> >
>
>
>
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