Forget this. I missed the single index on X.cum.mean[i]!!
On 3/12/07, jim holtman <[EMAIL PROTECTED]> wrote:
>
> I think you only have to calculate
>
> X.cum.sum[1:i]-seq(1:i)*X.cum.mean[i]
>
> once for the entire vector since it appears that you are just computing it
> over and over again for lengthing vectors. Here I have just computed it
> once and then did a running min/max on the data. This should run faster:
>
>
> system.time({
> X <- rnorm(10000)
> X.length <- length(X)
> X.cum.sum <- cumsum(X)
> X.cum.mean <- cumsum(X) # didn't have a cummean
> # create matrix of precalculated values (they are in the min/max columns)
> myMat <- cbind(newmean=X.cum.sum - (X.cum.mean * seq(X.length)), max=NA,
> min=NA)
> # now put min/max in columns; start with initial value
> myMat[1, c('max', 'min')] <- myMat[1, 'newmean']
> for (i in 2:nrow(myMat)){
> # determine the running min/max
> if (myMat[i, 'newmean'] > myMat[i-1, 'max']) myMat[i, 'max'] <-
> myMat[i, 'newmean']
> else myMat[i, 'max'] <- myMat[i-1, 'max']
> if (myMat[i, 'newmean'] < myMat[i-1, 'min']) myMat[i, 'min'] <-
> myMat[i, 'newmean']
> else myMat[i, 'min'] <- myMat[i-1, 'min']
> }
> })
>
>
>
>
>
> On 3/12/07, Oliver Faulhaber <[EMAIL PROTECTED]> wrote:
> >
> > Hi all,
> >
> > as I am trying to move slowly from just "working" to "good" code, I'd
> > like to ask if there's a smarter way than using a for-loop in tasks like
> > the example below.
> >
> > I need to obtain the extrema of the cumulated sum of a detrended time
> > series. The following code is currently used, please have a look at the
> > comments for my questions and remarks:
> >
> > system.time({
> > X <- rnorm(10000)
> > X.length <- length(X)
> > X.cum.sum <- cumsum(X)
> > X.cum.mean <- cummean(X)
> > # initializing the "output" vectors
> > X.min.detrended <- rep(NA,X.length)
> > X.max.detrended <- rep(NA,X.length)
> > for (i in 1:X.length) {
> > # Detrending of the time series from index 1 to i
> > # I think that's the time consuming part, are there any
> > # suggestions how to do this faster?
> > X.cum.sum.detrended <- X.cum.sum[1:i]-seq(1:i)*X.cum.mean[i]
> > # Calculating the min and max. Would a "range" be smarter here?
> > X.min.detrended[i] <- min( X.cum.sum.detrended)
> > X.max.detrended[i] <- max(X.cum.sum.detrended)
> > # As the programs takes rather long to complete I would like to
> > # get information about the progress. Any better way to do this
> > # than a cat(paste(i,"...",""))?
> > }
> > })
> >
> > As you can see this takes rather long even for this relative small
> > sample:
> >
> > [1] 41.11 0.97 45.14 NA NA
> >
> > I considered using "sapply", but expected problems with memory size as a
> >
> > lot more values have to be kept in memory (the typical length of X is
> > 10^6 - 10^7).
> >
> > Thanks again for your patience with newbies like me :)
> >
> > Regards
> > Oliver
> >
> > ______________________________________________
> > [email protected] mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide
> > http://www.R-project.org/posting-guide.html<http://www.r-project.org/posting-guide.html>
> > and provide commented, minimal, self-contained, reproducible code.
> >
>
>
>
> --
> Jim Holtman
> Cincinnati, OH
> +1 513 646 9390
>
> What is the problem you are trying to solve?
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem you are trying to solve?
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