timeseries - xst vs. dataframe?
On Feb 13, 2008 8:17 AM, icosa atropa <icos.atropa at gmail.com> wrote:
I've enjoyed your prosthelytizing of xts of late... the syntax you just highlighted looks appealing. I've found zoo powerful, and xts appears to be a useful extension. One question r.e. xts vs. dataframes: the lack of $a and [["a"]] notation in zoo has always struck me as a cumbersome difference from
zoo is modelled on the "ts" class, not on the "data.frame" class.
In R, the way it works is that $ is used on list-based objects
and not on array-based objects. Of course, your are free to define
and redefine operators as you please and since zoo is an S3 class
its possible to add your own S3 methods. $ indexing is less
than a dozen lines of code to add to your program:
"$.zoo" <- function(object, x) object[, x]
"$<-.zoo" <- function(object, x, value) {
stopifnot(length(dim(object)) == 2)
if (x %in% colnames(object)) object[,x] <- value
else {
object <- cbind(object, value)
colnames(object)[ncol(object)] <- x
}
object
}
# test
library(zoo)
z <- zoo(cbind(a = 1:3, b = 4:6))
z$c <- z$b + 1
z$a <- z$b - 1
z
dataframes. Is "list" syntax planned for inclusion in xst? At present, column numbering (test.zoo[,1]) seems the best alternative. Since as.data.frame(test.zoo)$a appears to recover the core data, it sounds sensible for "test.zoo$a" to extract an object containing the index and the named column. Does this break anything? e.g. : test.df = data.frame(a=1:5, b=2*(1:5)) test.df$a #[1] 1 2 3 4 5 index = Sys.time() + 60*1:5 test.zoo = zoo(test.df, order.by=index) test.zoo$a #NULL test.zoo[1:2,1] # 2008-02-13 05:54:40 2008-02-13 05:55:40 # 1 2 coredata(test.zoo)$a #NULL as.data.frame(test.zoo)$a # [1] 1 2 3 4 5 thanks and best, christian
At this point 'xts' objects behave much like any standard data.frame, matrix or, most closely, zoo object. They have some unique user 'xts' methods but all standard 'zoo' methods will work (it just extends 'zoo')