[R] Test of Presence Matrix HOWTO?

[Dimitris Rizopoulos]

you can use something like the following:

a <- list("A","B","C","D")
b <- list("A","B","E","F")
c <- list("A","C","E","G")

#####################

abc <- list(a, b, c)
unq.abc <- unique(unlist(abc))

out.lis <- lapply(abc, "%in%", x = unq.abc)
out.lis
lapply(out.lis, as.numeric)


I hope it helps.

Best,
Dimitris

----
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven

Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web: http://med.kuleuven.be/biostat/
     http://www.student.kuleuven.be/~m0390867/dimitris.htm


----- Original Message ----- 
From: "Johannes Graumann" <johannes_graumann at web.de>
To: <r-help at stat.math.ethz.ch>
Sent: Monday, February 26, 2007 4:25 PM
Subject: [R] Test of Presence Matrix HOWTO?


> Hello,
>
> Imagine 3 lists like so:
>
>> a <- list("A","B","C","D")
>> b <- list("A","B","E","F")
>> c <- list("A","C","E","G")
>
> What I need (vennDiagram) is a matrix characterizing with 1 or 0 
> whether any
> given member is present or not like so:
>     x1 x2 x3
> [1,]  1  1  1
> [2,]  1  1  0
> [3,]  1  0  1
> [4,]  1  0  0
> [5,]  0  1  1
> [6,]  0  1  0
> [7,]  0  0  1
>
> (where the rows represent "A"-"G" and the columns a-c, 
> respectively).
>
>> table(c(a,b,c))
> will give me a quick answer for the "1 1 1" case, but how to deal 
> with the
> other cases efficiently without looping over each string and looking 
> for
> membership %in% each list?
>
> Thanks for enlightening the learning,
>
> Joh
>
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