column transposition with xts

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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
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

? ? ? ?[[alternative HTML version deleted]]

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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
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

? ? ? ?[[alternative HTML version deleted]]

_______________________________________________
R-SIG-Finance at stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-sig-finance
-- Subscriber-posting only.
-- If you want to post, subscribe first.

--
Aleks Clark
zoo's lag function supports vector lags so try this
where x is your xts object:

z <- as.zoo(Cl(x))
zz <- lag(z, -seq(0, 20))

zz is a zoo object and you can use as.xts(zz) to convert it back to
xts if needed.

Note that lag.zoo follows the R convention of having negative lag mean
move the lagged series forward in time where lag.xts uses the opposite
convention.
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

? ? ? ?[[alternative HTML version deleted]]

_______________________________________________
R-SIG-Finance at stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-sig-finance
-- Subscriber-posting only.
-- If you want to post, subscribe first.

This probably won't be much better, but it is cleaner code:

*note: Lag is in quantmod*
tail(Lag(Cl(x),1:20))
Lag.1    Lag.2    Lag.3    Lag.4    Lag.5    Lag.6    Lag.7
2007-02-15 51.04699 50.90106 50.96653 50.91160 50.69562 50.67686 50.81383
2007-02-16 51.05185 51.04699 50.90106 50.96653 50.91160 50.69562 50.67686
2007-02-17 51.02164 51.05185 51.04699 50.90106 50.96653 50.91160 50.69562
2007-02-18 51.13653 51.02164 51.05185 51.04699 50.90106 50.96653 50.91160
2007-02-19 51.15151 51.13653 51.02164 51.05185 51.04699 50.90106 50.96653
2007-02-20 51.17899 51.15151 51.13653 51.02164 51.05185 51.04699 50.90106
              Lag.8    Lag.9   Lag.10   Lag.11   Lag.12   Lag.13   Lag.14
2007-02-15 50.60611 50.49865 50.69783 50.55509 50.43109 50.36928 50.35784
2007-02-16 50.81383 50.60611 50.49865 50.69783 50.55509 50.43109 50.36928
2007-02-17 50.67686 50.81383 50.60611 50.49865 50.69783 50.55509 50.43109
2007-02-18 50.69562 50.67686 50.81383 50.60611 50.49865 50.69783 50.55509
2007-02-19 50.91160 50.69562 50.67686 50.81383 50.60611 50.49865 50.69783
2007-02-20 50.96653 50.91160 50.69562 50.67686 50.81383 50.60611 50.49865
             Lag.15   Lag.16   Lag.17   Lag.18   Lag.19   Lag.20
2007-02-15 50.22578 50.02180 49.91875 49.88096 50.01091 50.07024
2007-02-16 50.35784 50.22578 50.02180 49.91875 49.88096 50.01091
2007-02-17 50.36928 50.35784 50.22578 50.02180 49.91875 49.88096
2007-02-18 50.43109 50.36928 50.35784 50.22578 50.02180 49.91875
2007-02-19 50.55509 50.43109 50.36928 50.35784 50.22578 50.02180
2007-02-20 50.69783 50.55509 50.43109 50.36928 50.35784 50.22578

HTH
Jeff
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

? ? ? ?[[alternative HTML version deleted]]

_______________________________________________
R-SIG-Finance at stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-sig-finance
-- Subscriber-posting only.
-- If you want to post, subscribe first.

--
Aleks Clark

_______________________________________________
R-SIG-Finance at stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-sig-finance
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-- If you want to post, subscribe first.

Jeffrey Ryan
jeffrey.ryan at insightalgo.com

ia: insight algorithmics
www.insightalgo.com