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
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 <motyocska at 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
Hi, library(stringr) b[str_detect(colnames(b),"^y")] ?# ?????? 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 #or b[,!is.na(match(gsub("\\..*","",names(b)),"y"))] #???????? 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 A.K. 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