3

This is really an extension of my question yesterday where I learned about apply.weekly. This works great, but I want to do this over wide zoo objects. If I use apply.weekly on a wide zoo it sums the columns, then performs the weekly aggregation:

> library(xts)
> set.seed(2001)
> zoo.daily <- zoo(data.frame(a=rnorm(20), b=rnorm(20), c=rnorm(20)), order.by=as.Date("2001-05-25") + 0:19)
> apply.weekly(zoo.daily, sum)
2001-05-27 2001-06-03 2001-06-10 2001-06-13 
  1.091999  -3.017688   3.842305   2.045370 
> apply.weekly(zoo.daily[, 1] + zoo.daily[, 2] + zoo.daily[, 3], sum) 
2001-05-27 2001-06-03 2001-06-10 2001-06-13 
  1.091999  -3.017688   3.842305   2.045370

I tried the apply family of operators, but those seem to strip out the zoo date index. I can do it in a for loop, but it's really time-consuming (much, much more than a factor of four slower than the aggregate function on as.yearmon periodicity). Here's the for loop:

week.ends <- index(zoo.daily[endpoints(zoo.daily, "weeks")[-1], ])
num.weeks <- nweeks(zoo.daily)
num.stocks <- ncol(zoo.daily)
zoo.weeks <- zoo(matrix(NA, num.weeks, num.stocks), order.by=week.ends)
for (i in seq(num.stocks)) {
    zoo.weeks[, i] <- apply.weekly(zoo.daily[, i], mean)
}

Which works (i.e., keeps each vector separate):

2001-05-27 -0.36663040 -0.108648725  0.8392788
2001-06-03  0.33032998  0.003025018 -0.7644534
2001-06-10  0.07816992  0.620198931 -0.1494681
2001-06-13  0.02114608  0.956226189 -0.2955824

Is there a way to quickly operate on all columns with apply.weekly? Thanks!

UPDATE: Joshua Ulrich points out that I need a column aware function (like colMeans or colSums). When I do this, I get the correct answers, but as a transposed matrix. Should I just reclass and move on? Or do I have an option/setting wrong?

> apply.weekly(zoo.daily, colSums)
        [,1]        [,2]       [,3]        [,4]
a -1.0998912  2.31230989  0.5471894  0.06343824
b -0.3259462  0.02117512  4.3413925  2.86867857
c  2.5178365 -5.35117351 -1.0462765 -0.88674717
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Richard Herron
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  • Regarding your update: install xts from R-forge. Jeff patched that a few weeks back, but it hasn't made it to CRAN yet. – Joshua Ulrich Nov 30 '10 at 21:54

1 Answers1

5

You need to use a column-aware function in apply.weekly. For example, use colSums instead of sum or colMeans instead of mean.

The more recent revisions of xts on R-forge give the output below. The version currently on CRAN returns the data transposed.

# install.packages("xts", repos="http://r-forge.r-project.org")
> apply.weekly(zoo.daily, colSums)
                     a           b          c
2001-05-27 -1.09989120 -0.32594617  2.5178365
2001-06-03  2.31230989  0.02117512 -5.3511735
2001-06-10  0.54718941  4.34139252 -1.0462765
2001-06-13  0.06343824  2.86867857 -0.8867472
> apply.weekly(zoo.daily, colMeans)
                     a            b          c
2001-05-27 -0.36663040 -0.108648725  0.8392788
2001-06-03  0.33032998  0.003025018 -0.7644534
2001-06-10  0.07816992  0.620198931 -0.1494681
2001-06-13  0.02114608  0.956226189 -0.2955824

If you need to use a custom function, you can use a combination of apply.weekly and apply:

> apply.weekly(zoo.daily, function(x) apply(x,2,mean))
                     a            b          c
2001-05-27 -0.36663040 -0.108648725  0.8392788
2001-06-03  0.33032998  0.003025018 -0.7644534
2001-06-10  0.07816992  0.620198931 -0.1494681
2001-06-13  0.02114608  0.956226189 -0.2955824
Joshua Ulrich
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