Thanks Rui and Ivan, works perfectly...
Andras
On Monday, February 4, 2019, 4:18:39 PM EST, Rui Barradas
<[email protected]> 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,
>
> 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'), by="day"), 3)
>
> d<-c(1,3,5)
> e<-c(1,4)
> f<-c(1,2)
>
> listA<-list(a,b,c)
> listB<-list(d,e,f)
>
>
> what I would like to do with a function (my real listA and listB can be of
> any length but always equal length, but their components like a,b,and c those
> can be unequal) as opposed to manually is to derive the following answer
>
> listfinal<-list(a[-d],b[-e],c[-f])
> listfinal
>
>
> essentially the elements in listB serve as identifying the position of
> corresponding list element in listA and removing it from listA.
>
> these lists listA and listB in practice are columns of a data frame that I am
> trying to work with and were generated with a function using lapply...
>
> appreciate any thoughts you may have to make this functional...
>
> thanks,
>
> Andras
>
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