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:

[hidden email]
[mailto:

[hidden email]] 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

[hidden email]
+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)])

>

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