Hi R users following this thread!

I evaluated both solutions given by Rui and Arun.
Both work very well.
Arun's is a little faster.

I did the following time measurements on a large matrix with 2 rows,
labelled "InterPro" and "GO", and 88664 columns, labelled with protein IDs.
I was interested in selecting those columns (proteins) that share a certain
"GO" annotation for molecular function.

So here is the time evaluation:
* Rui's method:
system.time(annos[ , sapply( annos[ "GO", ], function(x) { "GO:0005634"
%in% x } ), drop=F ] )
   user  system elapsed
  1.068   0.012   1.079
or
system.time(annos[ , unlist( lapply( annos[ "GO", ], function(x) {
"GO:0005634" %in% x } ) ), drop=F ])
   user  system elapsed
  0.876   0.000   0.880

* Arun's method:
system.time( annos[ , mapply( function(x){any(x=="GO:0005634")}, annos[
"GO", ] ), drop=F ] )
   user  system elapsed
  0.808   0.012   0.826

Thank you Rui and Arun very much for your help and everyone else for your
attention!
Kind regards!

2012/12/5 arun <smartpink...@yahoo.com>

> HI,
> Just tweaking my code also gives the same result:
> m[,mapply(function(x) any(x=="B"),m),drop=F]
> #     [,1]   [,2]
> #[1,] List,2 List,2
> A.K.
>
>
>
> ----- Original Message -----
> From: Asis Hallab <asis.hal...@gmail.com>
> To: Rui Barradas <ruipbarra...@sapo.pt>; r-help@r-project.org
> Cc:
> Sent: Tuesday, December 4, 2012 5:26 PM
> Subject: Re: [R] How to find matching columns in a matrix of lists?
>
> Hello,
>
> m[ , sapply(1:ncol(m), function(j) sapply("B", `%in%`, m[[1 , j]])), drop=F
> > ]
> >
>
> It indeed does.
>
> Thank you very much!
>
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>
> ______________________________________________
> R-help@r-project.org mailing list
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> PLEASE do read the posting guide
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>
>


-- 
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Skype: asis.hallab.cgn
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