[R] Counting occurances of a letter by a factor

[Davis, Brian]

Thomas,

I don't *believe* I have the heterozygote coded both ways.  However, I haven't check thoroughly.  I do notice that some SNPs (X's) only have 2 levels say AA and AT.  One could build the multiplication table from the levels I suppose.  (Or force the factor to include all three levels)

Brian

-----Original Message-----
From: Thomas Lumley [mailto:tlumley at u.washington.edu] 
Sent: Friday, September 10, 2010 3:22 PM
To: Davis, Brian
Cc: Phil Spector; r-help at r-project.org
Subject: Re: [R] Counting occurances of a letter by a factor

On Fri, 10 Sep 2010, Davis, Brian wrote:

> I'm my quest for brevity I think I scarified too much clarity.
>
> I'll try to be a little less brief in the hopes of being more clear.
>
> Say I have data frame like this as before:
>> DF<-data.frame(c("CC", "CC", NA, "CG", "GG", "GC"), c("L", "U", "L", "U", "L", NA))
>> colnames(DF)<-c("X", "Y")
>> DF
>     X    Y
> 1   CC    L
> 2   CC    U
> 3 <NA>    L
> 4   CG    U
> 5   GG    L
> 6   GC <NA>
>
> I need to count the frequency of the unique individual characters in DF$X at each factor level in DF$Y
>
> So for DF$Y == "L"  there are 2 "C"'s and 2 "G"'s
> and for DF$Y == "U" there are 3 "C"'s and 1 "G"
>
> The NA's should not contribute to the counts.
>
> If I had a individual character in DF$X instead of a string like:
>
>> DF2<-data.frame(c("C", "C", NA, "C", "G", "G"), c("L", "U", "L", "U", "L", NA))
>> colnames(DF2)<-c("X", "Y")
>> DF2
>     X    Y
> 1    C    L
> 2    C    U
> 3 <NA>    L
> 4    C    U
> 5    G    L
> 6    G <NA>
>
> Then table gives me exactly what I need.
>
>> table(DF2)
>   Y
> X   L U
>  C 1 2
>  G 1 0
>

I would use table() as the first step
> table(DF[,1],DF[,2])

      L U
   CC 1 1
   CG 0 1
   GC 0 0
   GG 1 0


and then multiply by a matrix that counts C and G:  
> cg<-rbind(C=c(2,1,1,0),G=c(0,1,1,2))
> cg%*%table(DF[,1],DF[,2])

     L U
   C 2 3
   G 2 1

If the genotype is a factor then you don't have to worry about empty genotypes.

Also, do you actually get the heterozygotes coded both ways?  When I have had to do this it has been simplified by having the heterozygotes all coded the same way (ie, only one of CG and GC appears), so that as.numeric() on the factor variable gives the number of copies of the alphabetically later allele.

          -thomas

>
> Hopefully this is a little bit clearer what I'm trying to accomplish.
>
> Brian
>
> -----Original Message-----
> From: Phil Spector [mailto:spector at stat.berkeley.edu]
> Sent: Friday, September 10, 2010 2:52 PM
> To: Davis, Brian
> Subject: Re: [R] Counting occurances of a letter by a factor
>
> Brian -
>    Here's the only thing I can come up with to give the
> same result as your "ans", but it doesn't seem to correspond
> with your description of the problem.
>
>> DF1 = DF
>> DF1$X = sapply(strsplit(as.character(DF$X),''),'[',1)
>> DF2 = DF
>> DF2$X = sapply(strsplit(as.character(DF$X),''),'[',2)
>> newDF = rbind(DF1,DF2)
>> table(newDF$Y,newDF$X)
>
>     C G
>   L 2 2
>   U 3 1
>
> 					- Phil Spector
> 					 Statistical Computing Facility
> 					 Department of Statistics
> 					 UC Berkeley
> 					 spector at stat.berkeley.edu
>
>
>
> On Fri, 10 Sep 2010, Davis, Brian wrote:
>
>> I'm trying to find a more elegant way of doing this.  What I'm trying to accomplish is to count the frequency of letters (major / minor alleles)  in  a string grouped by the factor levels in another column of my data frame.
>>
>> Ex.
>>> DF<-data.frame(c("CC", "CC", NA, "CG", "GG", "GC"), c("L", "U", "L", "U", "L", NA))
>>> colnames(DF)<-c("X", "Y")
>>> DF
>>     X    Y
>> 1   CC    L
>> 2   CC    U
>> 3 <NA>    L
>> 4   CG    U
>> 5   GG    L
>> 6   GC <NA>
>>
>> I have an ugly solution, which works if you know the factor levels of Y in advance.
>>
>>> ans<-rbind(table(unlist(strsplit(as.character(DF[DF[ ,'Y'] == 'L', 1]), ""))),
>> + table(unlist(strsplit(as.character(DF[DF[ ,'Y']  == 'U', 1]), ""))))
>>> rownames(ans)<-c("L", "U")
>>> ans
>>  C G
>> L 2 2
>> U 3 1
>>
>>
>> I've played with table, xtab, tabulate, aggregate, tapply, etc but haven't found a combination that gives a more general solution to this problem.
>>
>> Any ideas?
>>
>> Brian
>>
>> ______________________________________________
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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.
>>
>
> ______________________________________________
> 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.
>

Thomas Lumley
Professor of Biostatistics
University of Washington, Seattle