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

[R. Michael Weylandt]

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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