Sundar: As I understand it, you can easily create an index variable (a pointer, actually) that will pick out the y columns in order:
z<-yourdataframe y<-as.vector(z[,indexvar]) So if you could cbind() the x's, you'd be all set. Again, assuming I understand correctly, the x column you want is: x<-z[,-indexvar] ## still a frame/matrix nvec<-seq(length=ncol(x)) x<-as.vector(x[,rep(nvec,times=nvec)]) HTH -- and even if I got it wrong, it was fun, so thanks. -- Bert -- Bert Gunter Genentech Non-Clinical Statistics South San Francisco, CA "The business of the statistician is to catalyze the scientific learning process." - George E. P. Box > -----Original Message----- > From: [EMAIL PROTECTED] > [mailto:[EMAIL PROTECTED] On Behalf Of Sundar > Dorai-Raj > Sent: Tuesday, September 14, 2004 9:16 AM > To: R-help > Subject: [R] reshaping some data > > Hi all, > I have a data.frame with the following colnames pattern: > > x1 y11 x2 y21 y22 y23 x3 y31 y32 ... > > I.e. I have an x followed by a few y's. What I would like to > do is turn > this wide format into a tall format with two columns: "x", "y". The > structure is that xi needs to be associated with yij (e.g. x1 should > next to y11 and y12, x2 should be next to y21, y22, and y23, etc.). > > x y > x1 y11 > x2 y21 > x2 y22 > x2 y23 > x3 y31 > x3 y32 > ... > > I have looked at ?reshape but I didn't see how it could work > with this > structure. I have a solution using nested for loops (see below), but > it's slow and not very efficient. I would like to find a vectorised > solution that would achieve the same thing. > > Now, for an example: > > x <- data.frame(x1 = 1: 5, y11 = 1: 5, > x2 = 6:10, y21 = 6:10, y22 = 11:15, > x3 = 11:15, y31 = 16:20, > x4 = 16:20, y41 = 21:25, y42 = 26:30, y43 = 31:35) > # which are the x columns > nmx <- grep("^x", names(x)) > # which are the y columns > nmy <- grep("^y", names(x)) > # grab y values > y <- unlist(x[nmy]) > # reserve some space for the x's > z <- vector("numeric", length(y)) > # a loop counter > k <- 0 > n <- nrow(x) > seq.n <- seq(n) > # determine how many times to repeat the x's > repy <- diff(c(nmx, length(names(x)) + 1)) - 1 > for(i in seq(along = nmx)) { > for(j in seq(repy[i])) { > # store the x values in the appropriate z indices > z[seq.n + k * n] <- x[, nmx[i]] > # move to next block in z > k <- k + 1 > } > } > data.frame(x = z, y = y, row.names = NULL) > > ______________________________________________ > [EMAIL PROTECTED] mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide! > http://www.R-project.org/posting-guide.html >
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