[R] extracting column at regular intervals starting from different columns

Berend Hasselman bhh at xs4all.nl
Tue Sep 18 21:07:39 CEST 2012


On 18-09-2012, at 19:21, eliza botto wrote:

> 
> Dear useRs,
> i had a matrix with 31 rows and 444 columns and i wanted to extract every 37th column of that matrix starting from 1. more precisely i wanted to select columns 1, 38,75, 112 and so on. then doing the same by starting from column number 2(2,39,76,113.......).
> i was advised to use 
>> x[c(TRUE, rep(FALSE, 36)),]
> i works if it is start from first column but as i am very new to R i wanted to know what to do to make it work if i want to start column selection from column 2 and then from column 3 and so on.
> sorry for bothering you once again..

I have defined 2 more functions for David's and Clint's solution and my solution.

f3 <- function(x,M,start=1) {
    x[c(rep(FALSE,start-1),TRUE,rep(FALSE,M-start))]
} 

f4 <- function(x,M,start=1) {
    x[c(start,seq_len((length(x)-1)/M)*M+start)]
}

z3 <- f3(a,M,start=1)
z4 <- f4(a,M,start=1) 
identical(z3,z1)
identical(z3,z2)

z5 <- f3(a,M,start=4)
z6 <- f4(a,M,start=4)
identical(z5,z6)

With

N <- 444
a <- 1:N
M <- 37 

I get the following timing results.

> benchmark(f3(a,M,start=4), f4(a,M,start=4), replications=Nrep)
                  test replications elapsed relative user.self sys.self
 1 f3(a, M, start = 4)       100000   1.621    1.562     1.606    0.014
 2 f4(a, M, start = 4)       100000   1.038    1.000     1.035    0.002

Compared to the timings for start=1 (previous post) the solution with rep becomes slightly slower.

Berend




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