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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