From: William Dunlap <wdunlap_at_tibco.com>

Date: Tue, 26 Apr 2011 09:11:17 -0700

*> x <- matrix(runif(1e5*6), nrow=1e5)
*

> library(rbenchmark)

> benchmark(r1 <- f1(x), r2 <- f2(x), r3 <- f3(x), r4 <- f4(x),

+ replications=5, columns=c("test","replications","elapsed"), order="elapsed")

> identical(r1,r2) && identical(r1, r3) && identical(r1, r4)

**[1] TRUE
**
Bill Dunlap

Spotfire, TIBCO Software

wdunlap tibco.com

R-help_at_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. Received on Tue 26 Apr 2011 - 16:15:59 GMT

Date: Tue, 26 Apr 2011 09:11:17 -0700

A different approach is to use order() to sort first by row number and then break the ties by value. It is quick when there are lots of short rows.

> f1 <- function (x)

+ apply(x, 1, function(row) sort(row, decreasing = TRUE)[2])

> f2 <- function (x)

+ -apply(-x, 1, function(row) sort.int(row, partial = 2)[2])

> f3 <- function (x)

+ { + # order by row number then by value + y <- t(x) + array(y[order(col(y), y)], dim(y))[nrow(y) - 1, ] + }

> f4 <- function (x)

+ apply(x, 1, function(row) max(row[-which.max(row)]))

> library(rbenchmark)

> benchmark(r1 <- f1(x), r2 <- f2(x), r3 <- f3(x), r4 <- f4(x),

+ replications=5, columns=c("test","replications","elapsed"), order="elapsed")

test replications elapsed 3 r3 <- f3(x) 5 1.08 4 r4 <- f4(x) 5 12.59 2 r2 <- f2(x) 5 23.19 1 r1 <- f1(x) 5 59.54

> identical(r1,r2) && identical(r1, r3) && identical(r1, r4)

Spotfire, TIBCO Software

wdunlap tibco.com

> -----Original Message-----

*> From: r-help-bounces_at_r-project.org
**> [mailto:r-help-bounces_at_r-project.org] On Behalf Of peter dalgaard
**> Sent: Tuesday, April 26, 2011 8:13 AM
**> To: David Winsemius
**> Cc: r-help_at_r-project.org
**> Subject: Re: [R] Second largest element from each matrix row
**>
**>
**> On Apr 26, 2011, at 14:36 , David Winsemius wrote:
**>
**> >
**> > On Apr 26, 2011, at 8:01 AM, Lars Bishop wrote:
**> >
**> >> Hi,
**> >>
**> >> I need to extract the second largest element from each row of a
**> >> matrix. Below is my solution, but I think there should be
**> a more efficient
**> >> way to accomplish the same, or not?
**> >>
**> >>
**> >> set.seed(1)
**> >> a <- matrix(rnorm(9), 3 ,3)
**> >> sec.large <- as.vector(apply(a, 1, order, decreasing=T)[2,])
**> >> ans <- sapply(1:length(sec.large), function(i) a[i, sec.large[i]])
**> >> ans
**> >
**> > There are probably many but this one is reasonably compact,
**> one-step, and readable:
**> >
**> > > ans2 <- apply(a, 1, function(i) sort(i)[ dim(a)[2]-1])
**> > > ans2
**> >
**> > Refreshing my mail client proves I was right about many
**> solutions, but this is the first (so far) to use the dim attribute.
**>
**> Anything with sort() or order() will have complexity
**> O(n*log(n)) or worse (n is the number of columns), whereas
**> finding the k-th largest element has complexity O(k*n).
**>
**> For moderate n, this may be unimportant, but you could
**> potentially find a speedup using
**>
**> sort.int(i, decreasing=TRUE, partial=2)[2]
**>
**> or
**>
**> max(i[-which.max(i)])
**>
**> --
**> Peter Dalgaard
**> Center for Statistics, Copenhagen Business School
**> Solbjerg Plads 3, 2000 Frederiksberg, Denmark
**> Phone: (+45)38153501
**> Email: pd.mes_at_cbs.dk Priv: PDalgd_at_gmail.com
**>
**> ______________________________________________
**> R-help_at_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.
**>
*

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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. Received on Tue 26 Apr 2011 - 16:15:59 GMT

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