[R] List filtration

[Johannes Graumann]

Thanks for your help!

Joh

Dimitris Rizopoulos wrote:

> try this:
> 
> lis. <- lapply(lis, function(x) if (length(ind <- grep("^IPI", x)))
> x[ind[1]] else NULL)
> lis.[!sapply(lis., is.null)]
> 
> 
> 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: Thursday, February 22, 2007 3:33 PM
> Subject: [R] List filtration
> 
> 
>> Hello R-ologists,
>>
>> Imagine you have a list "list" like so:
>>
>>>list
>> [[1]]
>> [1] "IPI00776145.1" "IPI00776187.1"
>>
>> [[2]]
>> [1] "Something" "IPI00807764.1" "IPI00807887.1"
>>
>> [[3]]
>> [1] "IPI00807764.1"
>>
>> [[4]]
>> [1] "Somethingelse"
>>
>> What I need to achieve is a filtered list "list2" like so:
>>
>>>list2
>> [[1]]
>> [1] "IPI00776145.1"
>>
>> [[2]]
>> [1] "IPI00807764.1"
>>
>> [[3]]
>> [1] "IPI00807764.1"
>>
>> So:
>> - if sublist-entry 1 start with "^IPI" make it the list-entry.
>> - otherwise chose the first "^IPI" sublist-entry present.
>> - delete the list-entry if not "^IPI" sublist-entry present.
>>
>> Can anybody nudge me towards an elegant solution without looping - I
>> have
>> LOTS of entries to process ...
>>
>> Thanks for your Teachings,
>>
>> Joh
>>
>> ______________________________________________
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>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>> 
> 
> 
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> ______________________________________________
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> minimal, self-contained, reproducible code.