Forcing full weeks with apply.weekly()

问题

I'm trying to figure out what xts (or zoo) uses as the time after doing an apply.period. Consider the following:

> myTs = xts(1:10, as.Date(1:10, origin = '2012-12-1'))
> apply.weekly(myTs, colSums)
           [,1]
2012-12-02    1
2012-12-09   35
2012-12-11   19

I think the '2012-12-02' means "for the week ending 2012-12-02, the sum is 1". So basically the time is the end of the week.

But the problem is with that "2012-12-11" - I think what it's doing is saying that the 11th is the last day of the week that was given, so it's giving that as the time.

Is there any way to force it to give the sunday on which it ends, even if that day was not included in the data set?

回答1:

Try this:

nextsun <- function(x) 7 * ceiling(as.numeric(x-0+4) / 7) + as.Date(0-4)
aggregate(myTs, nextsun, sum)

where nextsun was derived from nextfri code given in the zoo quick reference by replacing 5 (for Friday) with 0 (for Sunday).

回答2:

Those are full weeks. It's only showing you the date of the very last observation. See ?endpoints (apply.weekly, is essentially a thin wrapper for endpoints).

apply.weekly
function (x, FUN, ...) 
{
    ep <- endpoints(x, "weeks")
    period.apply(x, ep, FUN, ...)
}
<environment: namespace:xts>

From ?endpoints

endpoints returns a numeric vector corresponding to the last observation in each period specified by on, with a zero added to the beginning of the vector, and the index of the last observation in x at the end.

Valid values for the argument on include: “us” (microseconds), “microseconds”, “ms” (milliseconds), “milliseconds”, “secs” (seconds), “seconds”, “mins” (minutes), “minutes”, “hours”, “days”, “weeks”, “months”, “quarters”, and “years”.

The answer to your second question is no, there is no option to do so. But you could always edit the last date manually, if you're going to present all data wrapped up anyways, I don't see any harm in it.

回答3:

No you can't force it give you the sunday.

Because the index of the result of period.apply is given by

ep <- endpoints(myTs,'weeks') 

myTs[ep]
           [,1]
2012-12-02    2
2012-12-09    9
2012-12-10   10

So you need to shift the last date. Unfortunately xts don't offer this option, you can't shift a single value of the index. I don't know why (maybe a design choice get unique index)

e.g You can do the flowing:

ts.weeks <- apply.weekly(myTs, colSums)
ts.weeks[length(ts.weeks)] <- last(index(myTs)) + 7-last(floor(diff(ep)))

来源：https://stackoverflow.com/questions/13944838/forcing-full-weeks-with-apply-weekly

标签

xts

zoo