R help - Oct 2007 - list matching

Dear list,

Given a list of elements like:
aa <- list(one=c("o", "n", "e"),
               tea=c("t", "e", "a"),
               thre=c("t", "h", "r",
"e"))

Is there a function that returns the intersection between all?
Both match() and intersect() only deal with two arguments, but sometimes I 
have more.

Here, it should return "e" (or NA if none of the letters are common).

I have a solution to apply %in% multiple times (here two times, first between 
the first two and then between the result and the third) but... perhaps there 
is a better and quicker way.

Thanks in advance,
Adrian

--
Adrian Dusa
Romanian Social Data Archive
1, Schitu Magureanu Bd
050025 Bucharest sector 5
Romania
Tel./Fax: +40 21 3126618 \
          +40 21 3120210 / int.101

On Sat, 2007-10-06 at 19:49 +0000, Adrian Dusa wrote:
> Dear list,
> 
> Given a list of elements like:
> aa <- list(one=c("o", "n", "e"),
>                tea=c("t", "e", "a"),
>                thre=c("t", "h", "r",
"e"))
> 
> Is there a function that returns the intersection between all?
> Both match() and intersect() only deal with two arguments, but sometimes I 
> have more.
> 
> Here, it should return "e" (or NA if none of the letters are
common).
> 
> I have a solution to apply %in% multiple times (here two times, first
between
> the first two and then between the result and the third) but... perhaps
there
> is a better and quicker way.
> 
> Thanks in advance,
> Adrian

intersectList <- function(x) 
{
  res <- table(unlist(sapply(x, unique)))
  names(res[res == length(x)])
}


In the first line, I use unique() to ensure that if the same letter
appears more than once in the same list element, it is not included in
the result set in error, such as in 'L' here:


L <- list(a = c("a", "b", "b"), b =
c("d", "b", "a"), c = c("d",
"a"))
> L
$a
[1] "a" "b" "b"

$b
[1] "d" "b" "a"

$c
[1] "d" "a"


So:
> intersectList(aa)
[1] "e"
> intersectList(L)
[1] "a"


HTH,

Marc Schwartz

zoo's merge can do a multiway intersection.  We turn each component
of aa into the times of a dataless zoo object (assuming the elements of
each component are unique) and merge them together using all = FALSE
which will only leave those points at times in all components.  Extracting
the time and stripping off the names gives the result.

library(zoo)
as.vector(time(do.call(merge, c(lapply(aa, function(x) zoo(,x)), all = FALSE))))


On 10/6/07, Adrian Dusa <dusa.adrian at gmail.com>
wrote:
>
> Dear list,
>
> Given a list of elements like:
> aa <- list(one=c("o", "n", "e"),
>               tea=c("t", "e", "a"),
>               thre=c("t", "h", "r",
"e"))
>
> Is there a function that returns the intersection between all?
> Both match() and intersect() only deal with two arguments, but sometimes I
> have more.
>
> Here, it should return "e" (or NA if none of the letters are
common).
>
> I have a solution to apply %in% multiple times (here two times, first
between
> the first two and then between the result and the third) but... perhaps
there
> is a better and quicker way.
>
> Thanks in advance,
> Adrian
>
> --
> Adrian Dusa
> Romanian Social Data Archive
> 1, Schitu Magureanu Bd
> 050025 Bucharest sector 5
> Romania
> Tel./Fax: +40 21 3126618 \
>          +40 21 3120210 / int.101
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

On 10/7/07, Adrian Dusa <dusa.adrian at gmail.com>
wrote:
> On Sunday 07 October 2007, Gabor Grothendieck wrote:
> > I thought the latest zoo fixes were on CRAN but perhaps they
> > are not.   Try this:
> >
> > library(zoo)
> > # next line loads latest version of merge.zoo (fixed in August)
> >
source("http://r-forge.r-project.org/plugins/scmsvn/viewcvs.php/*checkout*/
> >pkg/R/merge.zoo.R?rev=361&root=zoo")
> >
> > as.vector(time(do.call(merge, c(lapply(aa, function(x) zoo(,x)), all
> > FALSE))))
>
> Yeap, that fixes it.
> Ah-haa, so all = FALSE is merge()s argument...
> Neat :)
> Adrian
It also makes use of the fact that zoo is sufficiently general that the
times in the time series can be character strings.  For example, the
following time series has the value 1 at time "a", 2 at time
"b", etc.

zoo(1:26, letters)