[R] reshaping some data
Peter Wolf
s-plus at wiwi.uni-bielefeld.de
Tue Sep 14 19:07:59 CEST 2004
Try:
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)
df.names<-names(x)
ynames<-df.names[grep("y",df.names)]
xnames<-substring(sub("y","x",ynames),1,2)
cbind(unlist(x[,xnames]),unlist(x[,ynames]))
Peter
Sundar Dorai-Raj wrote:
> 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)
>
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