Hi,
May be this what you wanted.
res2 <- lapply(row.names(res[[1]]),function(x)
do.call(rbind,lapply(res,function(y) y[match(x, row.names(y)),])))
length(res2)
#[1] 48
dim(res2[[1]])
#[1] 2325 8
A.K.
On Monday, November 11, 2013 10:20 PM, Yu-yu Ren <renyan...@gmail.com> wrote:
Thank you so much for that script, it works great. One additional request; how
can I go about binding each of the 2325 matrices for each sample, resulting in
48 matrices of 8 column by 2325 row?
On Mon, Nov 11, 2013 at 10:02 PM, arun <smartpink...@yahoo.com> wrote:
>
>Hi,
>I already sent a reply to R-help. I am not sure about the "2342".
>
>set.seed(25)
>dat1 <-
>as.data.frame(matrix(sample(c("A","T","G","C"),46482*56,replace=TRUE),ncol=56,nrow=46482),stringsAsFactors=FALSE)
> lst1 <- split(dat1,as.character(gl(nrow(dat1),20,nrow(dat1))))
>res <- lapply(lst1,function(x) sapply(x[,1:8],function(y) sapply(x[,9:56],
>function(z) sum(y==z)/20)))
>
> length(res)
>#[1] 2325 ### check here
> dim(res[[1]])
>#[1] 48 8
>
>A.K.
>
>
>
>
>On Monday, November 11, 2013 10:00 PM, Yu-yu Ren <renyan...@gmail.com> wrote:
>
>Thank you, I have uploaded several example files, with intermediate outputs of
>what I have done and the logic flow.
>
>
>
>
>On Mon, Nov 11, 2013 at 9:37 PM, <smartpink...@yahoo.com> wrote:
>
>
>>Hi,
>>
>>Comparing the first 8 columns separately with 9-56 columns is not clear.
>>Also, please provide a reproducible example (using ?dput) for others to work
>>on.
>>
>>A.K.
>><quote author='Renyulb28'>
>>Hi all, I have a set of genetic SNP data that looks like
>>
>>Founder1 Founder2 Founder3 Founder4 Founder5 Founder6 Founder7 Founder8
>>Sample1 Sample2 Sample3 Sample...
>>A A A T T T T T A T A T
>>A A A T T T T T A T A T
>>A A A T T T T T A T A T
>>A A A T T T T T A T A T
>>A A A T T T T T A T A T
>>A A A T T T T T A T A T
>>A A A T T T T T A T A T
>>A A A T T T T T A T A T
>>A A A T T T T T A T A T
>>A A A T T T T T A T A T
>>A A A T T T T T A T A T
>>A A A T T T T T A T A T
>>
>>The size of the matrix is 56 columns by 46482 rows. I need to first bin the
>>matrix by every 20 rows, then compare each of the first 8 columns (founders)
>>to each columns 9-56, and divide the total number of matching
>>letters/alleles by the total number of rows (20). Ultimately I need 48 8
>>column by 2342 row matrices, which are essentially similarity matrices. I
>>have tried to extract each pair separately by something like
>>
>>"length(cbind(odd[,9],odd[,1])[cbind(odd[,9],cbind(odd[,9],odd[,1])[,1])[,1]=="T"
>>& cbind(odd[,9],odd[,1])[,2]=="T",])/nrow(cbind(odd[,9],odd[,1]))"
>>
>>but this is no where near efficient, and I do not know of a faster way of
>>applying the function to every 20 rows and across multiple pairs.
>>
>>In the example given above, if the rows were all identical like shown across
>>20 rows, then the first row of the matrix for Sample1 would be
>>
>>1 1 1 0 0 0 0
>>
>></quote>
>>Quoted from:
>>http://r.789695.n4.nabble.com/Apply-function-to-every-20-rows-between-pairs-of-columns-in-a-matrix-tp4680272.html
>>
>>
>>_____________________________________
>>Sent from http://r.789695.n4.nabble.com
>>
>>
>
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