[R] reshaping some data
Berton Gunter
gunter.berton at gene.com
Tue Sep 14 18:45:04 CEST 2004
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: r-help-bounces at stat.math.ethz.ch
> [mailto:r-help-bounces at stat.math.ethz.ch] 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)
>
> ______________________________________________
> R-help at 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
>
More information about the R-help
mailing list