Thanks Rui and Ivan, works perfectly...
Andras
On Monday, February 4, 2019, 4:18:39 PM EST, Rui Barradas
wrote:
Hello,
Like this?
Map('[', listA, lapply(listB, '*', -1))
Hope this helps,
Rui Barradas
Às 21:01 de 04/02/2019, Andras Farkas via R-help escreveu:
> Hello everyone,
>
>
Hello,
Like this?
Map('[', listA, lapply(listB, '*', -1))
Hope this helps,
Rui Barradas
Às 21:01 de 04/02/2019, Andras Farkas via R-help escreveu:
Hello everyone,
wonder if you would have a thought on a function for the following:
we have
a<-sample(seq(as.Date('1999/01/01'), as.Date('2
On Mon, 4 Feb 2019 21:01:06 + (UTC)
Andras Farkas via R-help wrote:
> listA<-list(a,b,c)
> listB<-list(d,e,f)
>
> what I would like to do with a function <...> as opposed to manually
> is to derive the following answer
>
> listfinal<-list(a[-d],b[-e],c[-f])
The `Map` function, unlike `lapp
Hello everyone,
wonder if you would have a thought on a function for the following:
we have
a<-sample(seq(as.Date('1999/01/01'), as.Date('2000/01/01'), by="day"),5)
b<-sample(seq(as.Date('1999/01/01'), as.Date('2000/01/01'), by="day"), 4)
c<-sample(seq(as.Date('1999/01/01'), as.Date('2000/01/01
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