Re: [R] help with data.frame

2013-05-21 Thread arun
Hi, library(stringr) b[str_detect(colnames(b),"^y")]  #    y   y.1   y.2 #1  0.0  0.00  0.00 #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

Re: [R] help with data.frame

2013-05-21 Thread Andras Farkas
that is exactly what I wanted! Thank you Sarah! Andras --- On Tue, 5/21/13, Sarah Goslee wrote: > From: Sarah Goslee > Subject: Re: [R] help with data.frame > To: "Andras Farkas" > Cc: r-help@r-project.org > Date: Tuesday, May 21, 2013, 2:07 PM > So if I understan

Re: [R] help with data.frame

2013-05-21 Thread Sarah Goslee
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 0.0 0.00 0.00 2 19.55811 17.023812 15.354880 3 10.74991

[R] help with data.frame

2013-05-21 Thread Andras Farkas
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