没有Rcpp可以加快速度吗？

Question

I'm looking to speed up the following algorithm. I give the function an xts time series and then want to perform a principal components analysis for each time point on the previous X points (I'm using 500 at the moment) and then use the results of that PCA (5 principal components in the following code) to compute some value. Something like this:

lookback <- 500
for(i in (lookback+1):nrow(x))
{   
        x.now <- x[(i-lookback):i]        
        x.prcomp <- prcomp(x.now)
        ans[i] <- (some R code on x.prcomp)
}

I assume this would require me to replicate the lookback rows as columns so that x would be something like cbind(x,lag(x),lag(x,k=2),lag(x,k=3)...lag(x,k=lookback)) , and then run prcomp on each line? This seems expensive though. Perhaps some variant of apply ? I'm willing to look into Rcpp but wanted to run this by you guys before that.

Edit: Wow thanks for all the responses. Info on my dataset/algorithm:

1. dim(x.xts) currently = 2000x24. But eventually, if this shows promise, it will have to run fast (I'll give it multiple datasets).
2. func(x.xts) takes ~70 seconds. That's 2000-500 prcomp calls with 1500 500x24 dataframe creations.

I attempted to use Rprof to see what was the most expensive part of the algo but it's my first time using Rprof so I need some more experience with this tool to get intelligible results (thanks for the suggestion).

I think I will first attempt to roll this into an _apply type loop, and then look at parallelizing.

Answer 1

On my 4 core desktop, if this wouldn't complete in a reasonable time-frame, I would run the chunk using something along the lines of (not tested):

library(snowfall)
sfInit(parallel = TRUE, cpus = 4, type = "SOCK")
lookback <- 500
sfExport(list = c("lookback", "x"))
sfLibrary(xts)

output.object <- sfSapply(x = (lookback+1):nrow(x),
    fun = function(i, my.object = x, lb = lookback) {
        x.now <- my.object[(i-lb):i]      
        x.prcomp <- prcomp(x.now)
        ans <- ("some R code on x.prcomp")

        return(ans)
    }, simplify = FALSE) # or maybe it's TRUE? depends on what ans is