an alternative is to use X2 below, which seems to be a little bit
faster:
N <- 1000
n <- 4
Ainv <- matrix(rnorm(N * N), N, N)
H <- matrix(rnorm(N * n), N, n)
d <- rnorm(N)
quad.form <- function (M, x){
jj <- crossprod(M, x)
return(drop(crossprod(jj, jj)))
}
#######################
#######################
invisible({gc(); gc(); gc()})
system.time(for(i in 1:200){
X0 <- solve(t(H) %*% Ainv %*% H) %*% t(H) %*% Ainv %*% d
}, gcFirst = TRUE)
invisible({gc(); gc(); gc()})
system.time(for(i in 1:200){
X1 <- solve(quad.form(Ainv, H), crossprod(crossprod(Ainv, H), d))
}, gcFirst = TRUE)
invisible({gc(); gc(); gc()})
system.time(for(i in 1:200){
tH.Ainv <- crossprod(H, Ainv)
X2 <- solve(tH.Ainv %*% H) %*% colSums(t(tH.Ainv) * d)
}, gcFirst = TRUE)
all.equal(X0, X1)
all.equal(X0, X2)
I hope this helps.
Best,
Dimitris
----
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web: http://www.med.kuleuven.be/biostat/
http://www.student.kuleuven.be/~m0390867/dimitris.htm
----- Original Message -----
From: "Robin Hankin" <[EMAIL PROTECTED]>
To: "Prof Brian Ripley" <[EMAIL PROTECTED]>
Cc: "RHelp" <r-help@stat.math.ethz.ch>; "Robin Hankin"
<[EMAIL PROTECTED]>
Sent: Wednesday, October 05, 2005 3:08 PM
Subject: Re: [R] eliminate t() and %*% using crossprod() and
solve(A,b)
>
> On 5 Oct 2005, at 12:15, Dimitris Rizopoulos wrote:
>
>> Hi Robin,
>>
>> I've been playing with your question, but I think these two lines
>> are not equivalent:
>>
>> N <- 1000
>> n <- 4
>> Ainv <- matrix(rnorm(N * N), N, N)
>> H <- matrix(rnorm(N * n), N, n)
>> d <- rnorm(N)
>> quad.form <- function (M, x){
>> jj <- crossprod(M, x)
>> return(drop(crossprod(jj, jj)))
>> }
>>
>>
>> X0 <- solve(t(H) %*% Ainv %*% H) %*% t(H) %*% Ainv %*% d
>> X1 <- solve(quad.form(Ainv, H), crossprod(crossprod(Ainv, H), d))
>> all.equal(X0, X1) # not TRUE
>>
>>
>> which is the one you want to compute?
>>
>
>
> These are not exactly equivalent, but:
>
>
> Ainv <- matrix(rnorm(1e6),1e3,1e3)
> H <- matrix(rnorm(4000),ncol=4)
> d <- rnorm(1000)
>
> X0 <- solve(t(H) %*% Ainv %*% H) %*% t(H) %*% Ainv %*% d
> X1 <- solve(quad.form(Ainv, H), crossprod(crossprod(Ainv, H), d))
> X0 - X1
> [,1]
> [1,] 4.884981e-15
> [2,] 3.663736e-15
> [3,] -5.107026e-15
> [4,] 5.717649e-15
>
> which is pretty close.
>
>
>
> On 5 Oct 2005, at 12:50, Prof Brian Ripley wrote:
>>
>>> QUESTION:
>>>
>>> how to calculate
>>>
>>> H %*% X
>>>
>>> in the recommended crossprod way? (I don't want to take a
>>> transpose
>>> because t() is expensive, and I know that %*% is slow).
>>>
>>
>> Have you some data to support your claims? Here I find (for random
>> matrices of the dimensions given on a machine with a fast BLAS)
>>
>>
>
> I couldn't supply any performance data because I couldn't figure out
> the
> correct R commands to calculate H %*% X without using %*% or t()!
>
> I was just wondering if there were a way to calculate
>
> H %*% solve(t(H) %*% Ainv %*% H) %*% t(H) %*% Ainv %*% d
>
> without using t() or %*%. And there doesn't seem to be (my original
> question didn't make it clear that I don't have X precalculated).
>
> My take-home lesson from Brian Ripley is that H %*% X is fast
> --but this is only useful to me if one has X precalculated, and in
> general I don't. But this discussion suggests to me that it might
> be
> a good idea to change my routines and calculate X anyway.
>
> thanks again Prof Ripley and Dimitris Rizopoulos
>
>
> very best wishes
>
> Robin
>
>
>
>>> system.time(for(i in 1:100) t(H) %*% Ainv)
>>>
>> [1] 2.19 0.01 2.21 0.00 0.00
>>
>>> system.time(for(i in 1:100) crossprod(H, Ainv))
>>>
>> [1] 1.33 0.00 1.33 0.00 0.00
>>
>> so each is quite fast and the difference is not great. However,
>> that is
>> not comparing %*% with crossprod, but t & %*% with crossprod.
>>
>> I get
>>
>>
>>> system.time(for(i in 1:1000) H %*% X)
>>>
>> [1] 0.05 0.01 0.06 0.00 0.00
>>
>> which is hardly 'slow' (60 us for %*%), especially compared to
>> forming X
>> in
>>
>>
>>> system.time({X = solve(t(H) %*% Ainv %*% H) %*% t(H) %*% Ainv %*%
>>> d})
>>>
>> [1] 0.04 0.00 0.04 0.00 0.00
>>
>> I would probably have written
>>
>>
>>> system.time({X <- solve(crossprod(H, Ainv %*% H), crossprod
>>> (crossprod(Ainv, H), d))})
>>>
>> 1] 0.03 0.00 0.03 0.00 0.00
>>
>> which is faster and does give the same answer.
>>
>> [BTW, I used 2.2.0-beta which defaults to gcFirst=TRUE.]
>>
>> --
>
>
>
>
>
> --
> Robin Hankin
> Uncertainty Analyst
> National Oceanography Centre, Southampton
> European Way, Southampton SO14 3ZH, UK
> tel 023-8059-7743
>
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