Re: [R] [FORGED] Re: Generate list if sequence form two vector element

Thanks everyone for the solutions !! Tanvir Ahamed Göteborg, Sweden | mashra...@yahoo.com - Original Message - From: Keith Jewell <keith.jew...@campdenbri.co.uk> To: r-h...@stat.math.ethz.ch Sent: Wednesday, 22 June 2016, 11:54 Subject: Re: [R] [FORGED] Re: Generate list if se

Re: [R] [FORGED] Re: Generate list if sequence form two vector element

or as a one-liner mapply(pmin(a, b), pmax(a,b), FUN=seq, SIMPLIFY=FALSE) On 22/06/2016 10:23, peter dalgaard wrote: There's also mapply(a, b, FUN=seq, SIMPLIFY=FALSE) (turn off simplication so that you don't unexpectedly get a matrix whenever all elements of results have same length. This

Re: [R] [FORGED] Re: Generate list if sequence form two vector element

There's also mapply(a, b, FUN=seq, SIMPLIFY=FALSE) (turn off simplication so that you don't unexpectedly get a matrix whenever all elements of results have same length. This also affects apply()-based solutions.) ...except that according to original spec, one should ensure a < b. So myseq <-

Re: [R] [FORGED] Re: Generate list if sequence form two vector element

Now why didn't I think of that? apply(matrix(c(a,b),ncol=2),1,function(x)x[1]:x[2]) Jim On Wed, Jun 22, 2016 at 6:14 PM, Rolf Turner wrote: > On 22/06/16 20:00, Jim Lemon wrote: >> >> Hi Tanvir, >> Not at all elegant, but: >> >> make.seq<-function(x)

Re: [R] [FORGED] Re: Generate list if sequence form two vector element

On 22/06/16 20:00, Jim Lemon wrote: Hi Tanvir, Not at all elegant, but: make.seq<-function(x) return(seq(x[1],x[2])) apply(matrix(c(a,b),ncol=2),1,make.seq) Not sure that this is more "elegant" but it's a one-liner: lapply(1:length(a),function(i,a,b){a[i]:b[i]},a=a,b=b) cheers, Rolf On