My comments have nothing to do with speed of your code,
but with the correctness.
> > np.lm <-function(dat, X, Y, ...){
> > # Ch 9.2: Slope est. (X) for Thiel statistic
> > ...
> > num[[i]] <- dat[j.s[i],Y] - dat[i.s[i],Y]
> > dom[[i]] <- dat[j.s[i],X] - dat[i.s[i],X]
Up to here is looks like X and Y are used to indicate which columns
of dat are the predictor and response, respectively.
> > ..
> > X <- median( sort( do.call(c, num) / do.call(c, dom) ) )
Now X is used to mean the slope of the regression.
> > # Ch 9.4: Intercept est. for Thiel statistic
> > Intercept <- median(dat[,"Y"] - X*dat[,"X"])
Now the predictor column must be named "X" and the response "Y",
no matter what your input X and Y were. Hence the function will fail
to give the correct answer in most cases.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -----Original Message-----
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf
> Of Berend Hasselman
> Sent: Sunday, October 21, 2012 11:59 AM
> To: Brad Schneid
> Cc: r-help@r-project.org
> Subject: Re: [R] help speeding up simple Theil regression function
>
>
> On 21-10-2012, at 20:06, Brad Schneid wrote:
>
> > Hello,
> >
> > I am working on a simple non-parametric (Theil) regression function and and
> > am following Hollander and Wolfe 1999 text. I would like some help making
> > my function faster. I have compared with pre-packaged version from "MBLM",
> > which isnt very fast either, but it appears mine is faster with N = 1000
> > (see results below). I plan on running this function repeatedly, and I
> > generally have data lengths of ~ N = 6000 or more.
> >
> > # My function following Hollander and Wolfe text, Chapter 9
> > np.lm <-function(dat, X, Y, ...){
> > # Ch 9.2: Slope est. (X) for Thiel statistic
> > combos <- combn(nrow(dat), 2)
> > i.s <- combos[1,]
> > j.s <- combos[2,]
> > num <- vector("list", length=length(i.s))
> > dom <- vector("list", length=length(i.s))
> >
> > for(i in 1:length(i.s)){
> > num[[i]] <- dat[j.s[i],Y] - dat[i.s[i],Y]
> > dom[[i]] <- dat[j.s[i],X] - dat[i.s[i],X]
> > }
> >
> > X <- median( sort( do.call(c, num) / do.call(c, dom) ) )
> > # Ch 9.4: Intercept est. for Thiel statistic
> > Intercept <- median(dat[,"Y"] - X*dat[,"X"])
> > out <- data.frame(Intercept, X)
> > return(out)
> > } # usage: np.lm(dat, X=1, Y=2)
> > ################################################################
> >
> > library("mblm") # I will compare to mblm() function
> >
> > X <- rnorm(1000)
> > Y <- rnorm(1000)
> > dat <- data.frame(X, Y)
> >
> > system.time(np.lm(dat, X=1, Y=2) )
> > user system elapsed
> > 118.610 0.130 119.144
> > 109.000 0.040 109.416 # ran it twice
> > 86.190 0.100 86.589 # 3rd time
>
> Alternative function without your i loop (it isn't needed and can be
> vectorized):
>
> np.lm.alt <-function(dat, X, Y, ...){
> # Ch 9.2: Slope est. (X) for Thiel statistic ==> (Pedantic comment: it
> is Theil (swap
> the i and e)
> combos <- combn(nrow(dat), 2)
> i.s <- combos[1,]
> j.s <- combos[2,]
>
> Y.num <- dat[j.s,Y] - dat[i.s,Y]
> X.dom <- dat[j.s,X] - dat[i.s,X]
> X <- median( Y.num / X.dom)
> # Ch 9.4: Intercept est. for Thiel statistic ==> (Pedantic comment: it
> is Theil (swap
> the i and e)
> Intercept <- median(dat[,"Y"] - X*dat[,"X"])
> out <- data.frame(Intercept, X)
> return(out)
> } # usage: np.lm(dat, X=1, Y=2)
>
>
> Try the compiler package on you original function:
>
> library(compiler)
> np.lm.c <- cmpfun(np.lm)
>
> Test speed and correct results:
>
> X <- rnorm(500)
> Y <- rnorm(500)
> dat <- data.frame(X, Y)
>
> system.time(npout.c <- np.lm.c(dat, X=1, Y=2) )
> system.time(npout.1 <- np.lm(dat, X=1, Y=2) )
> system.time(npout.a <- np.lm.alt(dat, X=1, Y=2) )
> identical(npout.1,npout.c)
> identical(npout.1,npout.a)
>
> Results:
>
> > system.time(npout.c <- np.lm.c(dat, X=1, Y=2) )
> user system elapsed
> 21.442 0.066 21.517
> > system.time(npout.1 <- np.lm(dat, X=1, Y=2) )
> user system elapsed
> 21.068 0.073 21.161
> > system.time(npout.a <- np.lm.alt(dat, X=1, Y=2) )
> user system elapsed
> 0.303 0.010 0.313
> > identical(npout.1,npout.c)
> [1] TRUE
> > identical(npout.1,npout.a)
> [1] TRUE
>
> You may try and test this with larger data lengths.
>
>
> Berend
>
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