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 also affects apply()-based
solutions.)
...except that according to original spec, one should ensure a < b. So
myseq <- function(a,b) if(a<b) a:b else b:a
mapply(a, b, FUN=myseq, SIMPLIFY=FALSE)
-pd
On 22 Jun 2016, at 10:42 , Jim Lemon <drjimle...@gmail.com> wrote:
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 <r.tur...@auckland.ac.nz> wrote:
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 Wed, Jun 22, 2016 at 5:32 PM, Mohammad Tanvir Ahamed via R-help
<r-help@r-project.org> wrote:
Hi,
I want to do the follow thing
Input :
a <- c(1,3,6,9)
b<-c(10,7,20,2)
Expected outcome :
d<-list(1:10,3:7,6:20,2:9)
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