column transposition with xts

tail(Lag(Cl(x),1:20))

Josh,

was hoping to avoid a for loop, but I guess this is as good as it gets
:) luckily I should only have to run this once.

thanks,

Aleks

On Mon, Sep 14, 2009 at 9:52 PM, Joshua Ulrich <josh.m.ulrich at gmail.com> wrote:

Aleks,

This isn't the most elegant solution, especially with large data sets,
but it is *a* solution. ?The code below yields a 24-column xts object
(20 lags of close prices, plus the original OHLC series).

library(xts)
library(quantmod)

data(sample_matrix)
x <- as.xts(sample_matrix[1:50,])

res <- x
for(i in 1:20) { res <- cbind(res,lag(Cl(x),i)) }

HTH,
Josh
--
http://www.fosstrading.com



On Mon, Sep 14, 2009 at 5:53 AM, Aleks Clark <aleks.clark at gmail.com> wrote:

I'm working with SVMs and have recently migrated to using TTR and thus xts,
but I've run into the following problem:

For each vector in an SVMs training set, I need to have a given amount of
historical data available as features for the vector. A simple example would
be to have the past 20 close prices. So I'd take a matrix like this:
? ?O H L C
1
2
3
4
5
...
N

and output something like this to give to my SVM:

O, H, L, C, C-1, C-2, C-3, etc
1
2
3
4
5

obviously at the top of the matrix, the historical Close cells would be
empty until there was enough data. In the pre-xts days, I accomplished this
using sapply() and t(), a probably hackish solution, but all I was able to
come up with. However, I am now running into problems with indices using
this technique, and was wondering if there was a better way to perform this
transformation, which really only amounts to a simple copy of data. Possible
solutions that occur to me include some sort of "index shift" and then a
merge, but I've been unable to get anything to work.


--
Aleks Clark

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--
Aleks Clark