On Fri Aug 21 09:12:03 1998,
Lars Karlsson <lars@ana.unibe.ch> wrote:
>Is there a way to avoid the loops below?
>vec is a vector of length n. I want to count the number of instances
>where an element and the next j-th element both equals 1.
...
>For example:
>> vec <- c(1,0,1,1,2,1,2,1,0,1)
>yields
>> aaa
> [1] 6 1 4 2 2 2 1 2 0 1
Hi,
I would work with the indices of the ones, as in this function:
> "spacing" <-
function(x, value = 1)
{
# DATE WRITTEN: 21 Aug 1998 LAST REVISED: 21 Aug 1998
# AUTHOR: Scott D. Chasalow (Scott.Chasalow@users.pv.wau.nl)
#
# DESCRIPTION:
# Count how many times x[j] and x[j + i] both equal value,
# j = 1,...,length(x); i = 0,...,length(x)-j.
#
# Returns an integer vector of length length(x), with one
# element for each value of i in 0,...,length(x)-1.
#
pos <- seq(along = x)[x == value]
if(length(pos) == 0)
return(rep(0, length(x)))
out <- outer(pos, pos, "-")
out <- out[lower.tri(out, diag = T)]
tabulate(out + 1, nbins = length(x))
}
> vec
[1] 1 0 1 1 2 1 2 1 0 1
> spacing(vec)
[1] 6 1 4 2 2 2 1 2 0 1
> spacing(vec, 0)
[1] 2 0 0 0 0 0 0 1 0 0
> spacing(vec, 2)
[1] 2 0 1 0 0 0 0 0 0 0
Cheers,
Scott
=========================================
Scott.Chasalow@users.pv.wau.nl
Wageningen Agricultural University
Laboratory of Plant Breeding
P.O. Box 386
6700 AJ Wageningen
THE NETHERLANDS
http://www.spg.wau.nl/pv/staff/Chasal_S.htm
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