[R] sample consecutive integers efficiently
Hi all,
I have some rough code to sample consecutive integers with length
according to a vector of lengths
#sample space (representing positions)
pos-c(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20)
#sample lengths
lengths-c(2,3,2)
From these two vectors I need a vector of sampled positions.
the sampling is without replacement, making things tough as the sampled
integers need to be consecutive. Im hoping somebody knows a faster way
of doing it than I have. ATM its way to slow on large vectors.
samplePos-function(l){
start.pos-sample(pos,1)
end.pos-start.pos+l-1
posies-start.pos:end.pos
posies
}
s.start-c()
newPos-function(a){
rp-samplePos(a)
#test sampled range is consecutive, if not resample
if (length(rp) != rp[a]+1 -rp[1]){rp-samplePos(a)}
pos-setdiff(pos,rp)
rp[1]
}
newps-c()
newps-unlist(lapply(lengths,newPos))
I think the bottleneck may be on the setdiff() function - the sample
space is quite large so I dont think there would be too many rejections.
Many thanks,
Chris
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Re: [R] sample consecutive integers efficiently
On Thu, 28 Aug 2008, Chris Oldmeadow wrote:
Hi all,
I have some rough code to sample consecutive integers with length according
to a vector of lengths
#sample space (representing positions)
pos-c(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20)
#sample lengths
lengths-c(2,3,2)
From these two vectors I need a vector of sampled positions.
the sampling is without replacement, making things tough as the sampled
integers need to be consecutive. Im hoping somebody knows a faster way of
doing it than I have. ATM its way to slow on large vectors.
samplePos-function(l){
start.pos-sample(pos,1)
end.pos-start.pos+l-1
posies-start.pos:end.pos
posies
}
s.start-c()
newPos-function(a){
rp-samplePos(a)
#test sampled range is consecutive, if not resample
if (length(rp) != rp[a]+1 -rp[1]){rp-samplePos(a)}
pos-setdiff(pos,rp)
rp[1]
}
newps-c()
newps-unlist(lapply(lengths,newPos))
I think the bottleneck may be on the setdiff() function - the sample space is
quite large so I dont think there would be too many rejections.
The bottleneck is in the formulation of the sampling scheme.
This is a simple combinatorics problem. There are 3360 possible values (
prod(16:14) ) for the start positions of the three elements, and you can
form a bijection between 1:3360 and the individual samples. If the number
of possible sample is small enough, it would be most efficient to sample
from the corresponding integer vector and then translate it to the
corresponding sample. For larger values where the number of possible
samples become a challenge for 32-bit integer arithmetic, I expect this
approach would be preferred:
Permute length ( pos ) - sum ( lengths ) + length( lengths ) distinct
(consecutively labelled) elements:
elz - sample( length ( pos ) - sum ( lengths ) + length( lengths ) )
Take the lengths of the original objects to be
z.lens - rep( 1, length( elz ) )
z.lens[ seq(along = lengths ) ] - lengths
(i.e. objects longer than 1 appear first)
Determine the start positions of the objects as if they were laid down
consecutively according to the permutation:
start - head( cumsum( c(0, z.lens[ elz ]) ) + 1 , -1 )
Find the start positions of just those with lengths greater than 1
gt.1 - match( seq(along=lengths) , elz )
Report the start positions
start[ gt.1 ]
---
If length( pos ) is large, you can rewrite the above to simply sample
the positions (in the ordering) of the objects with lengths greater than
1. You will have to revise the calculation of start and gt.1 in that case.
HTH,
Chuck
Many thanks,
Chris
__
R-help@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.
Charles C. Berry(858) 534-2098
Dept of Family/Preventive Medicine
E mailto:[EMAIL PROTECTED] UC San Diego
http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego 92093-0901
__
R-help@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.
Re: [R] sample consecutive integers efficiently
Charles C. Berry wrote:
On Thu, 28 Aug 2008, Chris Oldmeadow wrote:
Hi all,
I have some rough code to sample consecutive integers with length
according to a vector of lengths
#sample space (representing positions)
pos-c(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20)
#sample lengths
lengths-c(2,3,2)
From these two vectors I need a vector of sampled positions.
the sampling is without replacement, making things tough as the
sampled integers need to be consecutive. Im hoping somebody knows a
faster way of doing it than I have. ATM its way to slow on large
vectors.
samplePos-function(l){
start.pos-sample(pos,1)
end.pos-start.pos+l-1
posies-start.pos:end.pos
posies
}
s.start-c()
newPos-function(a){
rp-samplePos(a)
#test sampled range is consecutive, if not resample
if (length(rp) != rp[a]+1 -rp[1]){rp-samplePos(a)}
pos-setdiff(pos,rp)
rp[1]
}
newps-c()
newps-unlist(lapply(lengths,newPos))
I think the bottleneck may be on the setdiff() function - the sample
space is quite large so I dont think there would be too many rejections.
The bottleneck is in the formulation of the sampling scheme.
This is a simple combinatorics problem. There are 3360 possible values
( prod(16:14) ) for the start positions of the three elements, and you
can form a bijection between 1:3360 and the individual samples. If the
number of possible sample is small enough, it would be most efficient
to sample from the corresponding integer vector and then translate it
to the corresponding sample. For larger values where the number of
possible samples become a challenge for 32-bit integer arithmetic, I
expect this approach would be preferred:
Permute length ( pos ) - sum ( lengths ) + length( lengths ) distinct
(consecutively labelled) elements:
elz - sample( length ( pos ) - sum ( lengths ) + length( lengths ) )
Take the lengths of the original objects to be
z.lens - rep( 1, length( elz ) )
z.lens[ seq(along = lengths ) ] - lengths
(i.e. objects longer than 1 appear first)
Determine the start positions of the objects as if they were laid down
consecutively according to the permutation:
start - head( cumsum( c(0, z.lens[ elz ]) ) + 1 , -1 )
Find the start positions of just those with lengths greater than 1
gt.1 - match( seq(along=lengths) , elz )
Report the start positions
start[ gt.1 ]
---
If length( pos ) is large, you can rewrite the above to simply sample
the positions (in the ordering) of the objects with lengths greater
than 1. You will have to revise the calculation of start and gt.1 in
that case.
HTH,
Chuck
Many thanks,
Chris
__
R-help@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.
Charles C. Berry(858) 534-2098
Dept of Family/Preventive
Medicine
E mailto:[EMAIL PROTECTED]UC San Diego
http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego
92093-0901
Thanks! that worked perfectly.
Chris
__
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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.
