On Nov 4, 2010, at 4:24 PM, Sarah Goslee wrote:
Here's one possibility:
library(ecodist)
a <- c(1,1,1,2,2,3,3,3,3)
b <- c("a","b","c","a","d","a", "b", "e", "f")
x <- crosstab(a, b, rep(1, length(a)))
x
a b c d e f
1 1 1 1 0 0 0
2 1 0 0 1 0 0
3 1 1 0 0 1 1
x %*% t(x)
1 2 3
1 3 1 2
2 1 2 1
3 2 1 4
Antoher way:
> sapply(1:3, function(y) {
sapply(1:3, function(x){
length(intersect(b[a==y], b[a==x]) )
} ) } )
[,1] [,2] [,3]
[1,] 3 1 2
[2,] 1 2 1
[3,] 2 1 4
Sarah
On Thu, Nov 4, 2010 at 3:42 PM, cory n <corynis...@gmail.com> wrote:
Let's suppose I have userids and associated attributes... columns
a and b
a <- c(1,1,1,2,2,3,3,3,3)
b <- c("a","b","c","a","d","a", "b", "e", "f")
so a unique list of a would be
id <- unique(a)
I want a matrix like this...
[,1] [,2] [,3]
[1,] 3 1 2
[2,] 1 2 1
[3,] 2 1 4
Where element i,j is the number of items in b that id[i] and id[j]
share...
So for example, in element [1,3] of the result matrix, I want to see
2. That is, id's 1 and 3 share two common elements in b, namely "a"
and "b".
This is hard to articulate, so sorry for the terrible description
here. The way I have solved it is to do a double loop, looping over
every member of the id column and comparing it to every other member
of id to see how many elements of b they share. This takes forever.
Thanks
cn
--
Sarah Goslee
http://www.functionaldiversity.org
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