[R] Still help needed on embeded regression
Guojun Zhu
shmilylemon at yahoo.com
Wed May 3 04:02:42 CEST 2006
It does not work though. How is the lag work? How
does the lag work? I read the help and do not quite
understand. Here is a test
> y
[1] 1 2 3 4 5 6 7 8 9 10
> coredata(lag(y,-1))
[1] 1 2 3 4 5 6 7 8 9 10
attr(,"tsp")
[1] 2 11 1
--- Gabor Grothendieck <ggrothendieck at gmail.com>
wrote:
> Try
>
> runmean2 <- function(x, k) # k must be even
> (coredata(runmean(x, k-1)) * (k-1) +
> coredata(lag(x, -k/2, na.pad = TRUE)))/k
>
> Also, in your code use matrices or vectors instead
> of data frames
> to avoid any overhead in using data frames.
>
> On 5/2/06, Guojun Zhu <shmilylemon at yahoo.com> wrote:
> > Sorry to bother you guys again. This is great.
> But
> > this is for 61 number and the second case will
> change
> > 60 to 61. "run*" only accept odd number window.
> How
> > to get around it with 60? Any suggestion? Thanks.
> >
> > --- Gabor Grothendieck <ggrothendieck at gmail.com>
> > wrote:
> >
> > > Using runmean from caTools the first one below
> does
> > > it in under 1 second but will not handle NAs.
> The
> > > second one takes under 15 seconds and handles
> > > them by replacing them with linear
> approximations.
> > > Note that k must be odd.
> > >
> > > # 1
> > >
> > > library(caTools)
> > > set.seed(1)
> > > system.time({
> > > y <- rnorm(140001)
> > > x <- as.numeric(seq(y))
> > > k <- 61
> > > Mxy <- runmean(x * y, k)
> > > Mxx <- runmean(x * x, k)
> > > Mx <- runmean(x, k)
> > > My <- runmean(y, k)
> > > b <- (Mxy - Mx * My) / (Mxx - Mx * Mx)
> > > a <- My - b * Mx
> > > })
> > >
> > > # 2
> > >
> > > library(caTools)
> > > library(zoo)
> > > set.seed(1)
> > > system.time({
> > > y <- rnorm(140000)
> > > x <- as.numeric(seq(y))
> > > x[100:200] <- NA
> > > x <- na.approx(zoo(x))
> > > y <- zoo(y)
> > > k <- 60
> > > Mxy <- runmean(x * y, k)
> > > Mxx <- runmean(x * x, k)
> > > Mx <- runmean(x, k)
> > > My <- runmean(y, k)
> > > b <- (Mxy - Mx * My) / (Mxx - Mx * Mx)
> > > a <- My - b * Mx
> > > })
> > >
> > >
> > > On 5/1/06, Guojun Zhu <shmilylemon at yahoo.com>
> wrote:
> > > > I basically has a long data.frame a. but I
> only
> > > need
> > > > three columns x,y. Let us say the index of row
> is
> > > t.
> > > > I need to produce new column s_t as the linear
> > > > regression coefficient of
> (x_(t-60),...x_(t-1)) on
> > > > (y_(t-60),...,y_(t-1)). The data is about
> 140,000
> > > > rows. I wrote a simple code on this which is
> > > super
> > > > slow, it takes more than 2 hours on a 2.8Ghz
> Intel
> > > Duo
> > > > Core. My friend use SAS and his code needs
> only
> > > > couple of minutes. I know there must be some
> more
> > > > efficient way to write it. Can anyone help me
> on
> > > > this? Here is the code.
> > > >
> > > > Also one line produce a complete NA temp$y and
> lm
> > > > function failed on that. How to make it just
> > > produce
> > > > a NA instead and keep runing?
> > > >
> > > > attach(return)
> > > > betat=rep(NA,length(RET))
> > > > for (i in 61:length(RET)){cat(i," ");
> > > > if (year[[i]]>=1995){
> > > >
> > > >
> > >
> >
>
temp<-data.frame(y=RET[(i-60):(i-1)]-riskfree[(i-60):(i-1)],x=sprtrn[(i-60):(i-1)]-riskfree[(i-60):(i-1)])
> > > >
> > > >
> > >
> >
>
betat[[i]]<-lm(y~x+1,na.action=na.exclude,temp)[[1]][[2]]
> > > > #if (i%%100==0)
> > > > cat(i," ");
> > > >
> > > >
> > > >
> > >
> >
>
return$vol.cap[[i]]=mean(VOL[(i-12):(i-1)],na.rm=TRUE)/return$cap[[i]]
> > > > }
> > > > }
> > > >
> > > > ______________________________________________
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> > > >
> > >
> >
> >
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