[R] Taking specific/timed differences in a zoo timeseries

[Gabor Grothendieck]

Suggest you use chron class rather than Date class since the format
you indicate is the default format for chron.

> library(zoo)
> library(chron)
>
> set.seed(12345)
> data <- round(runif(27)*10+runif(27)*5, 0)
> dates <- chron(c("09/03/09", "09/04/09", "09/07/09", "09/09/09",
+ "09/10/09", "09/11/09", "09/14/09", "09/16/09", "09/17/09",
+ "09/18/09", "09/21/09", "09/22/09", "09/23/09",
+ "09/24/09", "09/25/09", "09/28/09", "09/29/09", "09/30/09",
+ "10/01/09", "10/02/09", "10/05/09", "10/06/09", "10/07/09",
+ "10/08/09", "10/09/09", "10/13/09", "10/14/09"))
> temp <- zoo(data, order.by=dates)
>
> rng <- range(time(temp))
> dt <- seq(rng[1], rng[2], "day")
> temp.m <- na.locf(merge(temp, zoo(, dt)))
>
> # create a lagged time scale and subtract the
> # lagged series from original
>
> dt.lag <- as.chron(as.yearmon(dt)+1/12) + as.numeric(month.day.year(dt)$day) - 1
> temp - zoo(coredata(temp.m), dt.lag)
10/05/09 10/06/09 10/07/09 10/08/09 10/09/09 10/13/09 10/14/09
      -5       -6        3        2       -2        2        1

See R News 4/1 for a relevant article and also its references.

On Wed, Oct 14, 2009 at 11:41 AM, Sergey Goriatchev <sergeyg at gmail.com> wrote:
> Dear Gabor,
>
> Thank you very much for your help!
> I'm now using your suggestion with my data.
>
> May I ask a stupid question?
> The output's index now has format "2009-10-14". How can I transform it back
> into original "10/14/09" and use this in a zoo object?
>
> Regards,
> Sergey
>
> On Wed, Oct 14, 2009 at 17:03, Gabor Grothendieck <ggrothendieck at gmail.com>
> wrote:
>>
>> Try this:
>>
>> library(zoo)
>> # temp <- ... from post asking question
>>
>> # create a day sequence, dt, with no missing days
>> # and create a 0 width series with those times.
>> # merge that with original series giving original
>> # series plus a bunch of times having NA values.
>> # Use na.locf to fill in those values with the last
>> # non-missing so far.
>>
>> rng <- range(time(temp))
>> dt <- seq(rng[1], rng[2], "day")
>>
>> temp.m <- na.locf(merge(temp, zoo(, dt)))
>>
>> # create a lagged time scale and subtract the
>> # lagged series from original
>>
>> dt.lag <- as.Date(as.yearmon(dt)+1/12) + as.numeric(format(dt, "%d")) - 1
>> temp - zoo(coredata(temp.m), dt.lag)
>>
>>
>> Using your data the output from the last line is:
>>
>> > temp - zoo(coredata(temp.m), dt.lag)
>> 2009-10-05 2009-10-06 2009-10-07 2009-10-08 2009-10-09 2009-10-13
>> 2009-10-14
>>        -5         -6          3          2         -2          2
>>  1
>>
>>
>> On Wed, Oct 14, 2009 at 10:39 AM, Sergey Goriatchev <sergeyg at gmail.com>
>> wrote:
>> > Hello everyone.
>> >
>> > I have a specific problem that I have difficulties to solve.
>> > Assume I have a zoo object:
>> >
>> > set.seed(12345)
>> > data <- round(runif(27)*10+runif(27)*5, 0)
>> > dates <- as.Date(c("09/03/09", "09/04/09", "09/07/09", "09/09/09",
>> > "09/10/09", "09/11/09", "09/14/09", "09/16/09", "09/17/09",
>> > "09/18/09", "09/21/09", "09/22/09", "09/23/09",
>> > "09/24/09", "09/25/09", "09/28/09", "09/29/09", "09/30/09",
>> > "10/01/09", "10/02/09", "10/05/09", "10/06/09", "10/07/09",
>> > "10/08/09", "10/09/09", "10/13/09", "10/14/09"), "%m/%d/%y")
>> > temp <- zoo(data, order.by=dates)
>> >
>> > What I need to do is to take differences between say October 14th and
>> > September 14, then October 13th and September 13th, that is 1 month
>> > difference independent of number of days inbetween. And when there is
>> > no matching date in an earlier month, like here where there is no
>> > September 13th, the date should be the first preceding date, that is
>> > September 11th in this example. How can I do that?
>> >
>> > The above is just an example, my zoo object is very big and I need to
>> > take differences between years, that is between October 14th, 2009 and
>> > October 14th, 2008, then Oct.13, 2009 and Oct.13, 2008, and so on.
>> > Also, the time index of my zoo object has format "10/14/09" (that is
>> > Oct.14, 2009), and that is the format I need to operate with and do
>> > not want to change. In the example I reformated just so that I can
>> > create a zoo object.
>> >
>> > Could some friendly person please show me how to do such a calculation?
>> >
>> > Thank you in advance!
>> >
>> > Best,
>> > Sergey
>> >
>> > ______________________________________________
>> > R-help at r-project.org mailing list
>> > https://stat.ethz.ch/mailman/listinfo/r-help
>> > PLEASE do read the posting guide
>> > http://www.R-project.org/posting-guide.html
>> > and provide commented, minimal, self-contained, reproducible code.
>> >
>
>
>
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