Message-ID: <AANLkTimPLRFFxexYtoGf5ezrvn3Uc9x8TH=2cEK7FPeX@mail.gmail.com>
Date: 2010-11-04T20:24:30Z
From: Sarah Goslee
Subject: avoid a loop
In-Reply-To: <AANLkTi=qOWXrp6YyTbORCiBjOHvBGxsxma4nZPRu2vJW@mail.gmail.com>
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
Sarah
On Thu, Nov 4, 2010 at 3:42 PM, cory n <corynissen at 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