A very similar question was asked on StackOverflow (by Mikhail? and then I guess
the answers there were somehow not satisfactory...)
http://stackoverflow.com/questions/16213029/more-efficient-strategy-for-which-or-match
where it turns out that a binary search (implemented in R) on the sorted vector
is much faster than sum, etc. I guess because it's log N without copying. The
more complicated condition x > .3 & x < .5 could be satisfied with multiple
calls to the search.
Martin
On 04/26/2013 01:20 PM, William Dunlap wrote:
I think the sum way is the best.
On my Linux machine running R-3.0.0 the sum way is slightly faster:
> x <- rexp(1e6, 2)
> system.time(for(i in 1:100)sum(x>.3 & x<.5))
user system elapsed
4.664 0.340 5.018
> system.time(for(i in 1:100)length(which(x>.3 & x<.5)))
user system elapsed
5.017 0.160 5.186
If you are doing many of these counts on the same dataset you
can save time by using functions like cut(), table(), ecdf(), and
findInterval(). E.g.,
system.time(r1 <- vapply(seq(0,1,by=1/128)[-1], function(i)sum(x>(i-1/128) &
x<=i), FUN.VALUE=0L))
user system elapsed
5.332 0.568 5.909
system.time(r2 <- table(cut(x, seq(0,1,by=1/128))))
user system elapsed
0.500 0.008 0.511
all.equal(as.vector(r1), as.vector(r2))
[1] TRUE
You should do the timings yourself, as the relative speeds will depend
on the version or dialect of the R interpreter and how it was compiled.
E.g., with the current development version of 'TIBCO Enterprise Runtime for R'
(aka 'TERR')
on this same 8-core Linux box the sum way is considerably faster then
the length(which) way:
> x <- rexp(1e6, 2)
> system.time(for(i in 1:100)sum(x>.3 & x<.5))
user system elapsed
1.87 0.03 0.48
> system.time(for(i in 1:100)length(which(x>.3 & x<.5)))
user system elapsed
3.21 0.04 0.83
> system.time(r1 <- vapply(seq(0,1,by=1/128)[-1], function(i)sum(x>(i-1/128) &
x<=i), FUN.VALUE=0L))
user system elapsed
2.19 0.04 0.56
> system.time(r2 <- table(cut(x, seq(0,1,by=1/128))))
user system elapsed
0.27 0.01 0.13
> all.equal(as.vector(r1), as.vector(r2))
[1] TRUE
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-----Original Message-----
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf
Of lcn
Sent: Friday, April 26, 2013 12:09 PM
To: Mikhail Umorin
Cc: r-help@r-project.org
Subject: Re: [R] speed of a vector operation question
I think the sum way is the best.
On Fri, Apr 26, 2013 at 9:12 AM, Mikhail Umorin <mike...@gmail.com> wrote:
Hello,
I am dealing with numeric vectors 10^5 to 10^6 elements long. The values
are
sorted (with duplicates) in the vector (v). I am obtaining the length of
vectors such as (v < c) or (v > c1 & v < c2), where c, c1, c2 are some
scalar
variables. What is the most efficient way to do this?
I am using sum(v < c) since TRUE's are 1's and FALSE's are 0's. This seems
to
me more efficient than length(which(v < c)), but, please, correct me if I'm
wrong. So, is there anything faster than what I already use?
I'm running R 2.14.2 on Linux kernel 3.4.34.
I appreciate your time,
Mikhail
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