[R] What is a better way to deal with lag/difference and loops in time series using R?

[R. Michael Weylandt]

Here's one:

x <- xts(1:5, Sys.Date() + 1:5)
y <- xts(rep(NA, length(x)), time(x))

or another, less direct but shorter:

y <- cbind(x, NA)[,2]

Michael

On Wed, Apr 11, 2012 at 4:16 AM, jpm miao <miaojpm at gmail.com> wrote:
> Hello,
>
>    Thank you very much.
>
>    How can I create an empty time series with the same dates as existing
> time series?
>
>    If x is a ts object, then it would be easy:
>    y<-ts(NA, start=start(x), end=end(x),frequency=frequency(x))
>
>    What can I do if x is a zoo or xts object?
>    I come up with a cumbersome way of doing this:
>    xts<-as.ts(x)
>    y1<-ts(NA, start=start(y), end=end(y),frequency=frequency(y))
>    y<-as.xts(y1)
>
>    Is there any easier way to do it?
>
>    Thanks
>
> miao
>
>
> 2012/4/11 R. Michael Weylandt <michael.weylandt at gmail.com>
>>
>> Two ways around this:
>>
>> I = Easy) Just use zoo/xts objects. ts objects a real pain in the
>> proverbial donkey because of things like this.
>>
>> Something like:
>>
>> library(xts)
>> PI1.yq <- as.xts(PI1) # Specialty class for quarterly data (or regular
>> zoo works)
>> lag(PI1.yq)
>>
>> II = Hard) lag on a ts actually changes the time indices while keeping
>> all the data, [so the fourth data point is the same value -- just a
>> different time point] what you may want to do is cbind() the objects
>> to see how they line up now.
>>
>> cbind(PI1, lag(PI1,4))
>>
>> Hope this helps,
>> Michael
>>
>> On Tue, Apr 10, 2012 at 11:21 PM, jpm miao <miaojpm at gmail.com> wrote:
>> > Hello,
>> >
>> >   I am writing codes for time series computation but encountering some
>> > problems
>> >
>> >   Given the quarterly data from 1983Q1 to 1984Q2
>> >
>> > PI1<-ts(c(2.747365190,2.791594762, -0.009953715, -0.015059485,
>> > -1.190061246, -0.553031799,  0.686874720,  0.953911035),
>> > start=c(1983,1), frequency=4)
>> >
>> >> PI1
>> >             Qtr1         Qtr2         Qtr3         Qtr4
>> > 1983  2.747365190  2.791594762 -0.009953715 -0.015059485
>> > 1984 -1.190061246 -0.553031799  0.686874720  0.953911035
>> >
>> >
>> >    If I would like to create a time series vector containing the data in
>> > 4
>> > quarters ahead
>> >
>> >> PI4<-lag(PI1,4)> PI4             Qtr1         Qtr2         Qtr3
>> >> Qtr4
>> > 1982  2.747365190  2.791594762 -0.009953715 -0.015059485
>> > 1983 -1.190061246 -0.553031799  0.686874720  0.953911035
>> >
>> >
>> >    Confusingly, PI1[1] and PI4[1] are exactly the same! I usually would
>> > like to calculate the difference between the vector of interest and  the
>> > corresponding values 4 quarters ahead, but it is zero!
>> >
>> >> PI1[1][1] 2.747365> PI4[1][1] 2.747365
>> >
>> >
>> > One remedy that comes into my mind is to use window,but a warning
>> > message
>> > emerges
>> >
>> >> PI4w<-window(PI4, start=start(PI1), end=end(PI1))Warning message:In
>> >> window.default(x, ...) : 'end' value not changed> PI4w           Qtr1
>> >> Qtr2       Qtr3       Qtr4
>> > 1983 -1.1900612 -0.5530318  0.6868747  0.9539110
>> >
>> >
>> >    Similar problems happen with the usage of the function "diff", which
>> > calculate the difference. I wonder if it is better to work with the
>> > dates
>> > (1983Q1, 1983Q2,.....) directly?
>> >
>> >    If I want to write a loop, say, to conduct some computation from
>> > 1983Q1
>> > to 2011Q4, the only way I know is to convert the dates to the ordinal
>> > indices, 1, 2, 3...... Can we work with the dates? Is there any built-in
>> > equality that provides the computation like
>> >    1983Q1 +1 equals 1983Q2?
>> >    In EViews, it is easy to do that. We can let %s run from 1983Q1 to
>> > 2011Q4, and he knows that 1983Q1+1 is exactly 1983Q2.
>> >
>> >     Thanks very much for your reply!
>> >
>> > miao
>> >
>> >        [[alternative HTML version deleted]]
>> >
>> > ______________________________________________
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>> > and provide commented, minimal, self-contained, reproducible code.
>
>