Re: [R] Looking for a more elegant solution than a loop

2015-10-26 Thread William Dunlap
> identical(as.list(x), xz) [1] TRUE Bill Dunlap TIBCO Software wdunlap tibco.com On Mon, Oct 26, 2015 at 6:31 PM, Erin Hodgess wrote: > Hello! > > The following (which is a toy example) works fine, but I wonder if there is > a better or more elegant way than to do the loop: > > xz <- vector("l

Re: [R] Looking for a more elegant solution than a loop

2015-10-26 Thread Erin Hodgess
t, > John > > - > John Fox, Professor > McMaster University > Hamilton, Ontario > Canada L8S 4M4 > Web: socserv.mcmaster.ca/jfox > > > > > -Original Message- > > From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of

Re: [R] Looking for a more elegant solution than a loop

2015-10-26 Thread Fox, John
t; From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Erin > Hodgess > Sent: October 26, 2015 9:32 PM > To: R help > Subject: [R] Looking for a more elegant solution than a loop > > Hello! > > The following (which is a toy example) works fine, but I wonder if there

[R] Looking for a more elegant solution than a loop

2015-10-26 Thread Erin Hodgess
Hello! The following (which is a toy example) works fine, but I wonder if there is a better or more elegant way than to do the loop: xz <- vector("list",length=4) x <- 6:9 for(i in 1:4)xz[[i]] <- x[i] xz [[1]] [1] 6 [[2]] [1] 7 [[3]] [1] 8 [[4]] [1] 9 This does exactly what I want, but th