On 08-Jan-10 21:09:08, MRKidd wrote:
> This will start off sounding very easy, but I think it will be
> very complicated.
>
> Let's say that I have a matrix, which shows the number of apples
> that each person in a group has.
>
> OriginalMatrix<-matrix(c(2,3,5,4,6),nrow=5,ncol=1,byrow=T,
> dimnames=list(c("Bob","Frank","Joe","Jim","David"),c("Apples")))
>
> Apples
> Bob 2
> Frank 3
> Joe 5
> Jim 4
> David 6
>
> I want to add a third column that shows what each person's rank is
> - e.g. David is number 1 because he has the most apples.
>
> So this is what I want:
> Apples Rank
> Bob 2 5
> Frank 3 4
> Joe 5 2
> Jim 4 3
> David 6 1
>
> I have managed to do this in the following steps:
> Unranked<-rownames(OriginalMatrix)
> Ranked<-names(sort(OriginalMatrix,decreasing=T))
> Matched<-match(Unranked,Ranked)
> NewMatrix<-cbind(OriginalMatrix,Matched)
>
> This is not acceptable, however, if two people have the same number of
> apples.
>
> You will get:
>
> NewMatrix
>
> Apples Rank
> Bob 2 5
> Frank 2 4
> Joe 5 2
> Jim 4 3
> David 6 1
>
> Does anyone know of a way to make it so that both Bob and Frank
> will be ranked as fourth (i.e.- tied for last place)?
The following is one way of approaching it (given in primitive form):
x<-matrix(c(2,2,5,4,6),ncol=1)
rownames(x)<-c("Bob","Frank","Joe","Jim","David")
x
# [,1]
# Bob 2
# Frank 2
# Joe 5
# Jim 4
# David 6
unique(sort(x))
# [1] 2 4 5 6
N <- length(x)
X <- cbind(x,rep(0,N))
Vals <- unique(sort(x))
M <- length(Vals)
for(i in (1:N)){X[X[,1]==Vals[i],2]<-(M+1-i)}
X
# [,1] [,2]
# Bob 2 4
# Frank 2 4
# Joe 5 2
# Jim 4 3
# David 6 1
Maybe this helps!
Ted.
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E-Mail: (Ted Harding) <ted.hard...@manchester.ac.uk>
Fax-to-email: +44 (0)870 094 0861
Date: 08-Jan-10 Time: 21:53:21
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