[R-SIG-Finance] Elegant bootstrapping with zoo

[Enrico Schumann]

hi brian,

one line of code, seems elegant to me ;) i guess sp500.subset has several
columns? i am not sure whether `sample' can be made to work row-wise on a
matrix, but you could write your own function.

require(zoo)

## create some artifical returns data
z <- rnorm(200) * 0.05
dim(z) <- c(100, 2)
z <- zoo(z, 1:100)

## function (x = matrix with time series in columns, nR = number of
replications)
bsRows <- function(x, nR){
	z  <- as.matrix(x)
	bs <- sample(seq(1,nrow(z)), nR, replace = TRUE)
	z  <- z[bs,]
	z  <- zoo(z, index(x)[bs])
	return(z)
}

## try
bsRows(z, 10)

(you may get warnings for repeated dates)
regards,
enrico

-----Ursprüngliche Nachricht-----
Von: r-sig-finance-bounces at stat.math.ethz.ch
[mailto:r-sig-finance-bounces at stat.math.ethz.ch] Im Auftrag von Rowe, Brian
Lee Yung (Portfolio Analytics)
Gesendet: Donnerstag, 26. März 2009 19:48
An: r-sig-finance at stat.math.ethz.ch
Betreff: [R-SIG-Finance] Elegant bootstrapping with zoo

Hello,

I want to bootstrap from a population in a zoo object. For example, I have a
set of returns for the S&P and want to bootstrap from this set over time. I
can do this with the below code, and I wanted to know if there is a more
elegant approach?

> h <- sp500.subset[sample(index(sp500.subset), 100, TRUE),]

I was hoping that a cleaner call would work, but apparently the length is
being used in the call to sample:

> h <- sample(sp500.subset, 100, TRUE)
Error in `[.zoo`(x, .Internal(sample(length(x), size, replace, prob))) :
  subscript out of bounds

Any thoughts or should I stick with the above approach?

Thanks,
Brian


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03/25/09
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