Dear list,
I have a long list of two vectors with some matching elements. I would like to
line them up in two columns and have an NA in those positions of the second
vector where a match is absent. With a simple example, I will explain my
problem.
(a<-1:6)
(b<-c(5,2))
(m1<-match(a,b))
(ab<-cbind(a,m1))
m2<-numeric(length(m1))
for (i in 1:length(m1))
{m2[i]<-ifelse(is.na(m1[i]),NA,b(m1[i]))}
# what I want to get - ab2 (shown below)
bsub<-c(NA,2,NA,NA,5,NA) # hoped to get this from m2 via the for loop above
(non NA elements from the b vector)
(ab2<-cbind(a,bsub))
I get an error message that the function b is not found. How do I define this
function?
Are there any other simpler methods for achieving my final result? Without a
loop?
I did find one solution that almost solved my problem.
aa<-cbind(a,a)
bb<-cbind(b,b)
abm<-merge(aa,bb,by.x="a",by.y="b",all=TRUE)
abm[,2:3] # just what I want
However, if I choose a little more complicated version of the problem :
rm(list=ls())
(a<-c(8,5,4,7,3,4,5,2,1))
(b<-c(5,2))
(a1<-sort(a))
(b1<-sort(b))
aa<-cbind(a1,a1)
bb<-cbind(b1,b1)
(ab<-merge(aa,bb,by.x="a1",by.y="b1",all=TRUE))
ab[,2:3]
In this case, there is a match for both the 5's in a1. I want only one match as
there is only one 5 in b.
My question is - is there any generalised solution for the above problem?
One extra feature that would be interesting is if there are elements in the 2nd
vector that are missing in the 1st.
In that case, I would like to have NA's in the 1st vector matching the extra
elements in the 2nd vector.
I will appreciate any help that I can get.
Thanks,
Ravi
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