R help - Apr 2012 - What is a better way to deal with lag/difference and loops in time series using R?

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]]

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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> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.

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)
>
> ? ?If I would like to create a time series vector containing the data in 4
> quarters ahead
>
> ? ?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.
>
This takes the difference between each point and the point 4 quarters
back and then lags the result forward.

lag(diff(PI1, 4), 4)

In zoo you can do this:

library(zoo)

z <- as.zoo(PI1)
time(z) <- as.yearqtr(time(z))

# quarter after '83 Q2
z[as.yearqtr("1983 Q2")+1/4]

You might also want to look at rollapply in the zoo package.

xts which is related to zoo has flexible subscripting.

If your data is regular as in your example then you could also use the
tis package which is related to the fame package but can be used
without it.

zoo has several vignettes which give many examples of processing time series.

The timeSeries package is another alternative.

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