Jeffrey J. Hallman frb.gov> writes:
> Tim Hesterberg insightful.com> writes:
>
> > toby_marks americancentury.com asked:
> > >I am trying to divide the columns of a matrix by the first row in the
> > >matrix.
> >
> > Dividing columns of a matrix by a vector is a pretty fundamental
> > operati
Tim Hesterberg <[EMAIL PROTECTED]> writes:
> [EMAIL PROTECTED] asked:
> >I am trying to divide the columns of a matrix by the first row in the
> >matrix.
>
> Dividing columns of a matrix by a vector is a pretty fundamental
> operation, and the query resulted in a large number of suggestions:
>
[EMAIL PROTECTED] asked:
>I am trying to divide the columns of a matrix by the first row in the
>matrix.
Dividing columns of a matrix by a vector is a pretty fundamental
operation, and the query resulted in a large number of suggestions:
x/matrix(v, nrow(x), ncol(x), byrow = TRUE))
sweep(x, 2, v
What version of R was this?
In 2.4.0 alpha
> a <- matrix(1:24,4)
> system.time(for(i in 1:1000) junk <- a / rep(a[1,], each = 4))
[1] 0.014 0.000 0.014 0.000 0.000
> system.time(for(i in 1:1000) junk <- t(t(a)/a[1,]))
[1] 0.057 0.000 0.058 0.000 0.000
shows a large margin the other way, which in
In terms of speed Toby's original idea was actually the fastest.
Here they are decreasing order of the largest timing in each
row of system.time. I also tried it with a 100x10 matrix and
got almost the same order:
> library(reshape)
> system.time(for(i in 1:1000) iapply(a, 1, "/", a[1,]))
[1] 11.
The apply was exactly what I was after. And, I will check out the others
as well. great tips!
"Gabor Grothendieck" <[EMAIL PROTECTED]>
09/06/2006 11:11 AM
To
"[EMAIL PROTECTED]" <[EMAIL PROTECTED]>
cc
r-help@stat.math.ethz.ch
Subject
Re: [R] Matrix
Prof. Brian Ripley wrote:
> On Wed, 6 Sep 2006, Christos Hatzis wrote:
>
> > See ?sweep
> >
> > sweep(a, 2, a[1,],"/")
>
> That is less efficient than
>
> a/rep(a[1,], each=nrow(a))
*My* first instinct was to use
t(t(a)/a[1,])
(which has not heretofore been suggested).
This seems t
Yet another one using the idempotent apply in reshape package
that eliminates the transpose:
library(reshape)
iapply(a, 1, "/", a[1,])
On 9/6/06, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> This last one could also be written slightly shorter as:
>
> t(apply(a, 1, "/", a[1,]))
>
> On 9/6/06,
This last one could also be written slightly shorter as:
t(apply(a, 1, "/", a[1,]))
On 9/6/06, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> And here is one more:
>
> t(apply(a, 1, function(x) x/a[1,]))
>
> On 9/6/06, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> > Here are a few possibilitie
> [EMAIL PROTECTED]
> Sent: Wednesday, September 06, 2006 11:49 AM
> To: r-help@stat.math.ethz.ch
> Subject: [R] Matrix multiplication using apply() or lappy() ?
>
> I am trying to divide the columns of a matrix by the first row in the
> matrix.
>
> I have tried to get
And here is one more:
t(apply(a, 1, function(x) x/a[1,]))
On 9/6/06, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> Here are a few possibilities:
>
> a <- matrix(1:24, 4) # test data
>
> a / rep(a[1,], each = 4)
>
> a / outer(rep(1, nrow(a)), a[1,])
>
> a %*% diag(1/a[1,])
>
> sweep(a, 2, a[1,],
Here are a few possibilities:
a <- matrix(1:24, 4) # test data
a / rep(a[1,], each = 4)
a / outer(rep(1, nrow(a)), a[1,])
a %*% diag(1/a[1,])
sweep(a, 2, a[1,], "/")
On 9/6/06, [EMAIL PROTECTED]
<[EMAIL PROTECTED]> wrote:
> I am trying to divide the columns of a matrix by the first row in th
See ?sweep
sweep(a, 2, a[1,],"/")
-Christos
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of
[EMAIL PROTECTED]
Sent: Wednesday, September 06, 2006 11:49 AM
To: r-help@stat.math.ethz.ch
Subject: [R] Matrix multiplication using apply() or lappy
I am trying to divide the columns of a matrix by the first row in the
matrix.
I have tried to get this using apply and I seem to be missing a concept
regarding the apply w/o calling a function but rather command args %*% /
etc. Would using apply be more efficient than this approach?
I have o
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