that is exactly what I wanted! Thank you Sarah!
Andras
--- On Tue, 5/21/13, Sarah Goslee <sarah.gos...@gmail.com> wrote:
> From: Sarah Goslee <sarah.gos...@gmail.com>
> Subject: Re: [R] help with data.frame
> To: "Andras Farkas" <motyoc...@yahoo.com>
> Cc: r-help@r-project.org
> Date: Tuesday, May 21, 2013, 2:07 PM
> 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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