Use a list to store the partial matrices: z <- vector("list", 189)
for(i in 1:189) z[[i]] <- subset(...) -Christos -----Original Message----- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of antonio rodriguez Sent: Tuesday, November 07, 2006 2:01 PM To: R-Help Subject: [R] subsetting a matrix and filling other Hi, Having a matrix F(189,6575) I want to do this: z1<-subset(F[,1], F[,1] >= 5 & F[,1] <= 10) . . . z189<-subset(F[,189], F[,189] >= 5 & F[,189] <= 10) I would prefer to have an empty matrix, say 'z' in order to fill its columns with the output of subsetting F. But each of the subsets can differ in length. It's to say: length(z1) 1189 length(z2) 1238 Don't know how to set up this with a for loop or the use of apply: z<-189 for (i in 1:189) {z[i]<-subset(F.zoo[,i], F.zoo[,i] >= 5 & z1 <= 10)} Error in z[i] <- subset(F.zoo[, i], F.zoo[, i] >= 5 & z1 <= 10) : nothing to replace with or z<-matrix(0,6575,189) for (i in 1:189) {z[i]<-subset(F.zoo[,i], F.zoo[,i] >= 5 & z1 <= 10)} new error Thanks, Antonio ______________________________________________ R-help@stat.math.ethz.ch 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. ______________________________________________ R-help@stat.math.ethz.ch 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.