[R] list with list function

[Rui Barradas]

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
> 
> ______________________________________________
> R-help using r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>