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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