[R] correlation and matrix

[Jacques VESLOT]

it should be smth like that:

apply(sapply(seq(1, 204, by=12), seq, length=4), 2, function(x)
 {
 M <- dta[,x]
 z <- sapply(M, nlevels) # if dta is a dataframe
 if (sum(z==1)<3) cor(as.matrix(M[,z!=0]), use="comp", method="spear") 
else NA
 })  

Jacques VESLOT

INRA - Biostatistique & Processus Spatiaux
Site Agroparc 84914 Avignon Cedex 9, France

Tel: +33 (0) 4 32 72 21 58
Fax: +33 (0) 4 32 72 21 84



Nathalie.Cornileau at csiro.au a écrit :
> Dear everyone,
>  
> I am new in R and I've got difficulties in realizing the following
> tasks:
> -I have variables (factors) with different numbers of levels, either 1,
> 2 or 3.
> -I have a matrix containing these 204 factors and I have to correlate
> them by groups of 4 variables.
> -I have to delete the factors just having one level ( because when
> correlating one-level factors, the output is NA)
>  
> here is my code:
> lst<-seq(1, 204, by=12) % there are 12 factors for 17 natural resources
> for (n in lst)
> { 
> Mx<- matrix(0, byrow = F, ncol = 4, nrow=nrow(dta)) % I extract the 4
> factors I have to correlate and I'd like to do it for each n
> {if (nlevels(dta[,n+4])!=1) 
> Mx[,1]<-dta[,n+4]
> else
> Mx[,1]<-NA}
> {if (nlevels(dta[,n+5])!=1) 
> Mx[,2]<-dta[,n+5]
> else
> Mx[,2]<-NA}
> {if (nlevels(dta[,n+7])!=1) 
> Mx[,3]<-dta[,n+7]
> else
> Mx[,3]<-NA}
> {if (nlevels(dta[,n+8])!=1) 
> Mx[,4]<-dta[,n+8]
> else
> Mx[,4]<-NA}
> p<-0                % I compute the number of non - NA columns and I'd
> like to delete the Na columns from that matrix
>  
> for (i in 1:4)
> {
> if(!is.na(sum(Mx[,i])>0)) p<-p+1   
> }
> print(p)
> {if (p==0 | p==1) stop("computation impossible")
>   else {
>   r<-0
>   for (i in 1:4)
> {
> if(is.na(sum(Mx[,i])>0))  r<-i
> }
> print(r)
> print(cor((as.matrix(Mx[,-r])), use="complete.obs", method="spearman"))
> }
> }
> } %The problem is the last step doesn't work for p==2.
>  In fact, it seems the loop for doesn't work either.
>  
> I hope it is clear enough and I thank you in advance for your help.
> Nathalie
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