Re: [R] quadratic form
If you meant QR vs. inverting X'X for linear regression, the motivation for
using QR is not speed, but numerical stability. There's no univerally good
least squares algorithm that would be uniformly better than anything else
for any kind of data.
Andy
> From: Jari Oksanen
>
>
> On 3 Nov 2005, at 17:25, Robin Hankin wrote:
>
> > Hi Alvarez
> >
> >
> > If you define
> >
> > quad.form.inv <- function (M, x)
> > {
> > drop(crossprod(x, solve(M, x)))
> > }
> >
> > then you will avoid an expensive call to %*% as well.
> >
> Is %*% really expensive in all platforms? I had a function
> that used QR
> decomposition instead of quadratic forms, but then I got a
> message from
> Canada suggesting that %*% would be faster. Indeed, it was in
> not-too-large data sets and in Mac (or powerpc). I run some
> tests with
> real applications, and found that my 800MHz iBook G4 run like
> a 2.5GHz
> Intel machine when %*% was used. This really was architecture
> dependent, since the performance boost was similar under OS X
> and Linux
> in the very same PowerPC. So it seems that %*% is very cheap if you
> have PowerPC, but it may be expensive in Intel. (I also run a test in
> Sun, and it was somewhere between Intel and PowerPC.)
>
> cheers, jari oksanen
> --
> Jari Oksanen, Oulu, Finland
>
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Re: [R] quadratic form
Peter Dalgaard <[EMAIL PROTECTED]> writes:
> Alvarez Pedro <[EMAIL PROTECTED]> writes:
>
> > On page 22 of the R-introduction guide it's written:
> >
> > the quadratic form x^{'} A^{-1} x which is used in
> > multivariate computations, should be computed by
> > something like x%*%solve(A,x), rather than computing
> > the inverse of A.
> >
> > Why isn't it good to compute t(x) %*% solve(A) %*% x?
>
> It's just a waste of CPU time. For k x k matrices, solution of Ax=b is
> of computational complexity O(k^2) whereas inversion of A is O(k^3).
> This is obviously more important for k=1000 than for k=5.
Erm, Thomas Lumley points out that solution of Ax=b is also of order
k^3 ingeneral (R uses a QR decomposition). So it's just the multiplier
that differs. The savings should still be on the order of k^3 though.
--
O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B
c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
(*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918
~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907
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Re: [R] quadratic form
On 3 Nov 2005, at 17:25, Robin Hankin wrote:
> Hi Alvarez
>
>
> If you define
>
> quad.form.inv <- function (M, x)
> {
> drop(crossprod(x, solve(M, x)))
> }
>
> then you will avoid an expensive call to %*% as well.
>
Is %*% really expensive in all platforms? I had a function that used QR
decomposition instead of quadratic forms, but then I got a message from
Canada suggesting that %*% would be faster. Indeed, it was in
not-too-large data sets and in Mac (or powerpc). I run some tests with
real applications, and found that my 800MHz iBook G4 run like a 2.5GHz
Intel machine when %*% was used. This really was architecture
dependent, since the performance boost was similar under OS X and Linux
in the very same PowerPC. So it seems that %*% is very cheap if you
have PowerPC, but it may be expensive in Intel. (I also run a test in
Sun, and it was somewhere between Intel and PowerPC.)
cheers, jari oksanen
--
Jari Oksanen, Oulu, Finland
__
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Re: [R] quadratic form
Hi Alvarez
If you define
quad.form.inv <- function (M, x)
{
drop(crossprod(x, solve(M, x)))
}
then you will avoid an expensive call to %*% as well.
HTH
Robin
On 3 Nov 2005, at 13:01, Alvarez Pedro wrote:
> On page 22 of the R-introduction guide it's written:
>
> the quadratic form x^{'} A^{-1} x which is used in
> multivariate computations, should be computed by
> something like x%*%solve(A,x), rather than computing
> the inverse of A.
>
> Why isn't it good to compute t(x) %*% solve(A) %*% x?
>
> Thanks a lot for help!
>
> __
> R-help@stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide! http://www.R-project.org/posting-
> guide.html
>
--
Robin Hankin
Uncertainty Analyst
National Oceanography Centre, Southampton
European Way, Southampton SO14 3ZH, UK
tel 023-8059-7743
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Re: [R] quadratic form
Alvarez Pedro <[EMAIL PROTECTED]> writes:
> On page 22 of the R-introduction guide it's written:
>
> the quadratic form x^{'} A^{-1} x which is used in
> multivariate computations, should be computed by
> something like x%*%solve(A,x), rather than computing
> the inverse of A.
>
> Why isn't it good to compute t(x) %*% solve(A) %*% x?
It's just a waste of CPU time. For k x k matrices, solution of Ax=b is
of computational complexity O(k^2) whereas inversion of A is O(k^3).
This is obviously more important for k=1000 than for k=5.
--
O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B
c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
(*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918
~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907
__
R-help@stat.math.ethz.ch mailing list
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Re: [R] quadratic form
On Thu, 3 Nov 2005, Alvarez Pedro wrote:
> On page 22 of the R-introduction guide it's written:
>
> the quadratic form x^{'} A^{-1} x which is used in
> multivariate computations, should be computed by
> something like x%*%solve(A,x), rather than computing
> the inverse of A.
>
> Why isn't it good to compute t(x) %*% solve(A) %*% x?
The answer is only two lines above;
Numerically, it is both inefficient and potentially unstable to compute
@code{x <- solve(A) %*% b} instead of @code{solve(A,b)}.
See also the footnote.
--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
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Re: [R] quadratic form
Alvarez Pedro wrote:
> On page 22 of the R-introduction guide it's written:
>
> the quadratic form x^{'} A^{-1} x which is used in
> multivariate computations, should be computed by
> something like x%*%solve(A,x), rather than computing
> the inverse of A.
>
> Why isn't it good to compute t(x) %*% solve(A) %*% x?
>
> Thanks a lot for help!
>
> __
> R-help@stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
See
Bates, D. (2004): Least Squares Calculations in R. R News 4(1), 17-20,
http://CRAN.R-project.org/doc/Rnews/
Uwe Ligges
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Re: [R] quadratic form
Pedro:
Solving the equation in this fashion can be computationally slow and
unstable. There is R an article in R News that answers this question
directly and is a pretty easy read with good examples. Check it out at
the following link:
http://cran.r-project.org/doc/Rnews/Rnews_2004-1.pdf
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Alvarez Pedro
Sent: Thursday, November 03, 2005 8:02 AM
To: r-help@stat.math.ethz.ch
Subject: [R] quadratic form
On page 22 of the R-introduction guide it's written:
the quadratic form x^{'} A^{-1} x which is used in multivariate
computations, should be computed by something like x%*%solve(A,x),
rather than computing the inverse of A.
Why isn't it good to compute t(x) %*% solve(A) %*% x?
Thanks a lot for help!
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide!
http://www.R-project.org/posting-guide.html
__
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[R] quadratic form
On page 22 of the R-introduction guide it's written:
the quadratic form x^{'} A^{-1} x which is used in
multivariate computations, should be computed by
something like x%*%solve(A,x), rather than computing
the inverse of A.
Why isn't it good to compute t(x) %*% solve(A) %*% x?
Thanks a lot for help!
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
