On Feb 5, 2010, at 1:50 AM, Bert Gunter wrote:
Folks:
You can make use of matrix subscripting and avoid R level loops and
applys
altogether. This will end up being many times faster.
Here's your original code:
Z=matrix(rnorm(20), nrow=4)
index=replicate(4, sample(1:5, 3))
P=4
tmpr=list()
for (i in 1:P)
{
tmp = Z[i,index[,i]]
tmpr[[i]]=tmp
}
for clarity, here's the index matrix I got:
index
[,1] [,2] [,3] [,4]
[1,] 5 1 2 3
[2,] 2 2 4 4
[3,] 1 5 5 5
Here's what I got for tmpr when I used your code:
tmpr
[[1]]
[1] -0.6246316 -0.8695538 -0.4136176
[[2]]
[1] 0.02885345 -1.89837071 0.43195955
[[3]]
[1] 0.2453368 -0.1788287 -0.6620405
[[4]]
[1] -0.87077697 -1.62554371 0.04464793
So the ith component of tmpr is is just what the indices in the ith
column
of index pick out of the ith row of Z. That is, the first component
of tmpr
are the (1,5), (1,2), and (1,1) elements of Z. Matrix (in general,
array)indexing -- read the man page for "[" carefully: it's
documented in
the "Matrices and Arrays" section -- allow you to "stack" these
pairs (for
n-dim arrays,n-tuples) row-wise into a matrix and use this matrix as
an
index:
Z[cbind(c(1,1,1),index[,1])]
[1] -0.6246316 -0.8695538 -0.4136176
So you can do everything at once by (making use of R's columnwise
storage of
arrays) as:
result <- Z[cbind(as.vector(col(index)), as.vector(index))]
which gives:
[1] -0.62463163 -0.86955383 -0.41361765 0.02885345 -1.89837071
0.43195955
0.24533679
[8] -0.17882867 -0.66204048 -0.87077697 -1.62554371 0.04464793
Note that this vector is the same as: unlist(tmpr). So you can turn
it into
a matrix e.g. where column i is the ith component of tmpr by:
dim(result) <- dim(index)
As I said, for large problems, this should be wayyyyy faster than
explicit
loops or the hidden (and optimized, but still) loops of apply
functions.
Well, twice as fast as the explicit anyway:
> system.time( replicate(10000, {result <-
Z[cbind(as.vector(col(index)), as.vector(index))]; dim(result) <-
dim(index)} )
+ )
user system elapsed
0.164 0.001 0.171
> system.time( replicate(10000, for (i in 1:P) { tmpr[[i]] <-
Z[i,index[,i]] } ) )
user system elapsed
0.267 0.049 0.330
Which was in turn twice as fast as the lapply approach:
> system.time( replicate(10000, tmpr[1:4]<-lapply(1:4, function(i, x,
y) {x[i,y[,1]]}, Z, index ) ) )
user system elapsed
0.628 0.015 0.646
--
David.
Bert Gunter
Genentech Nonclinical Statistics
-----Original Message-----
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org
] On
Behalf Of RICHARD M. HEIBERGER
Sent: Thursday, February 04, 2010 9:10 PM
To: Carrie Li
Cc: r-help
Subject: Re: [R] How do I use "tapply" in this case ?
lapply(1:4, function(i, x, y) {x[i,y[,1]]}, Z, index ) ## reproduces
your results
sapply(1:4, function(i, x, y) {x[i,y[,1]]}, Z, index ) ## collapses
your list into a set of columns
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
