Hi,
You could also use:
library(seqinr)
x1<- rep(c('A','G','C','T'),4)
splitseq(x1,frame=0,word=2)
#[1] "AG" "CT" "AG" "CT" "AG" "CT" "AG" "CT"
splitseq(x1,frame=1,word=2)
#[1] "GC" "TA" "GC" "TA" "GC" "TA" "GC"
splitseq(x1,frame=0,word=3)
#[1] "AGC" "TAG" "CTA" "GCT" "AGC"
A.K.
----- Original Message -----
From: R. Michael Weylandt <michael.weyla...@gmail.com>
To: Abhishek Pratap <abhishek....@gmail.com>
Cc: "r-help@r-project.org" <r-help@r-project.org>
Sent: Thursday, April 4, 2013 9:14 PM
Subject: Re: [R] about subsetting vectors/list in R
On Thu, Apr 4, 2013 at 7:55 PM, Abhishek Pratap <abhishek....@gmail.com> wrote:
> On Thu, Apr 4, 2013 at 5:53 PM, R. Michael Weylandt
> <michael.weyla...@gmail.com> wrote:
>> by_two <- function(x, collapse = ""){
>> dim(x) <- c(length(x) / 2, 2)
>> apply(x, 1, function(y) paste(y, collapse = collapse))
>> }
>
> Thanks.. just wondering if this will be slick for list/vectors with 100
> thousands of entries. ?
No, the apply() loop likely isn't optimal. But I can do
x <- rep(letters, length.out = 1e6)
system.time(by_two(x)) # Approx 15 seconds
on my slow old machine so this might be one of those cases of "good
enough and come worry about it if profiling shows its a real bottle
neck later".
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and provide commented, minimal, self-contained, reproducible code.
