So if I understand you correctly, and I may not, you want to extract
the columns from a dataframe that start with y?
Using your reproducible example (thanks!):
> b[, grepl("^y", colnames(b))]
y y.1 y.2
1 0.00000 0.000000 0.000000
2 19.55811 17.023812 15.354880
3 10.74991 9.024250 8.177128
4 5.91924 4.789331 4.367188
Sarah
On Tue, May 21, 2013 at 2:01 PM, Andras Farkas <motyoc...@yahoo.com> wrote:
> Dear All
>
> I have the following code for list "a":
>
> a <-list(structure(c(0, 4, 8, 12, 0, 19.5581076131386, 10.7499105081144,
> 5.91923975728553, 0, 4.08916328337685, 2.26872955281708, 1.24929641535359
> ), .Dim = c(4L, 3L), .Dimnames = list(NULL, c("time", "y", "b"
> )), istate = c(2L, 107L, 250L, NA, 5L, 5L, 0L, 52L, 22L, NA,
> NA, NA, NA, 0L, 1L, 1L, NA, NA, NA, NA, NA), rstate = c(0.867511261090201,
> 0.867511261090201, 12.7772879103809, 0, 0), lengthvar = 2L, class =
> c("deSolve",
> "matrix"), type = "lsoda"), structure(c(0, 4, 8, 12, 0, 17.0238115689622,
> 9.02425032330714, 4.7893314106951, 0, 4.45067278743554, 2.37140075611636,
> 1.25855947034654), .Dim = c(4L, 3L), .Dimnames = list(NULL, c("time",
> "y", "b")), istate = c(2L, 106L, 251L, NA, 4L, 4L, 0L, 52L, 22L,
> NA, NA, NA, NA, 0L, 1L, 1L, NA, NA, NA, NA, NA), rstate = c(0.662055762167652,
> 0.662055762167652, 12.3096826617166, 0, 0), lengthvar = 2L, class =
> c("deSolve",
> "matrix"), type = "lsoda"), structure(c(0, 4, 8, 12, 0, 15.3548797334796,
> 8.17712839316703, 4.36718847853436, 0, 5.15624657530424, 2.77411694866808,
> 1.48166036763212), .Dim = c(4L, 3L), .Dimnames = list(NULL, c("time",
> "y", "b")), istate = c(2L, 108L, 260L, NA, 5L, 5L, 0L, 52L, 22L,
> NA, NA, NA, NA, 0L, 1L, 1L, NA, NA, NA, NA, NA), rstate = c(0.735884123193699,
> 0.735884123193699, 12.1878866053931, 0, 0), lengthvar = 2L, class =
> c("deSolve",
> "matrix"), type = "lsoda"))
>
> then I convert it to "b"
>
> b <-data.frame(a)
>
> and manually I would extract the y variables (y, y.1 and y.2) as follows
>
> d <-t(cbind(b$y,b$y.1,b$y.2))
>
> Currently I only have 3 y variables, so manual solution is very easy. I would
> like to ask if you have any thoughts on how I could "automate" (or extract
> all ys) this so that I could achieve the same goal even if I have 5000 ys
> (from y, y.1, y.2.... to y.5000) or any other number of ys for that matter
> with a simple code (as opposed to something like d
> <-t(cbind(b$y,b$y.1,b$y.2,....b$y.5000))).
>
> your help is greatly appreciated,
>
> thanks,
>
> Andras
>
--
Sarah Goslee
http://www.functionaldiversity.org
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