# [R] finding common elements in a list

Berton Gunter gunter.berton at gene.com
Sat Apr 8 23:38:50 CEST 2006

```The original post is ambiguous: do you want to find the intersection or do
you want to find whether a prespecified set is in the intersection? Patrick
provided you an answer to the latter while you provided an answer to the
former. Actually, I thought using table as you did (mod the need for no
replicates) was clever. A more direct but I think considerably slower
approach would be to use intersect() in a loop:

inall<-intersect(foo[[1]],foo[[,2]])
for(i in seq(3, to=length(foo))inall<-intersect(inall,foo[[i]])

I suspect you already thought of this and rejected it. Other than
transparency, I think the only advantage it has is that it will work for
something other than lists of numerics, e.g. it will work for lists of
factors, which the table() solution would not.

-- Bert

-----Original Message-----
From: r-help-bounces at stat.math.ethz.ch
[mailto:r-help-bounces at stat.math.ethz.ch] On Behalf Of Patrick Burns
Sent: Friday, April 07, 2006 12:04 PM
To: Andy Bunn
Cc: R-Help
Subject: Re: [R] finding common elements in a list

Here is one solution:

> all(unlist(lapply(foo, function(x) c(2,3) %in% x)))
[1] TRUE

This doesn't have the restriction of assuming that the components
of the list have unique elements, as the original solution does.

Patrick Burns
patrick at burns-stat.com
+44 (0)20 8525 0696
http://www.burns-stat.com
(home of S Poetry and "A Guide for the Unwilling S User")

Andy Bunn wrote:

>Suppose I have a list where I want to extract only the elements that occur
>in every component. For instance in the list foo I want to know that the
>numbers 2 and 3 occur in every component. The solution I have seems
>unnecessarily clunky. TIA, Andy
>
>    foo <- list(x = 1:10, y=2:11, z=1:3)
>    bar <-unlist(foo)
>    bartab <- table(bar)
>    as.numeric(names(bartab)[bartab==length(foo)])
>
>	[[alternative HTML version deleted]]
>
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