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R-USERS-L  2014

R-USERS-L 2014

Subject:

Re: Need help on Rolling standard deviation with jump

From:

"Bhattacharya,Souvik" <[log in to unmask]>

Reply-To:

UF R Users List <[log in to unmask]>

Date:

Sun, 31 Aug 2014 12:05:20 -0400

Content-Type:

text/plain

Parts/Attachments:

Parts/Attachments

text/plain (104 lines)

Hi Forrest,

Thanks for your suggestions. But this rolling sd computes the sd for 
the rows, 1:5, 6:11, 12:17,....What I want is to calculate the sd for 
the rows 1:5, 2:6, 3:7, likewise
Thanks
Souvik




On Sun, 31 Aug 2014 10:25:23 -0400, Forrest Stevens wrote:
> I think that rollapply() is actually what you want unless I'm
> misunderstanding your data structure. For example, this is what I 
> think
> you're trying to accomplish:
>
>
> require(xts)
>
> d <- zooreg(rnorm(50), order.by=as.Date(1:50), deltat=1/365)
>
> ## Make an irregular time series out of it:
> di <- sample(d, 40)
>
> ## This I think approximates the data you have:
> ds <- xts(di)
> dz <- merge(ds, d)
>
> s <- rollapply(dz[,1], width=5, by=5, FUN=sd, na.rm=TRUE)
>
>
> It would help if you supplied a full sample of your data and the code
> you're using to prepare it, or at least something like the above.  
> But
> hopefully this gets you part of the way there.
>
> Sincerely,
> Forrest
>
> --
> Forrest R. Stevens
> Ph.D. Candidate, QSE3 IGERT Fellow
> Department of Geography
> Land Use and Environmental Change Institute
> University of Florida
> www.clas.ufl.edu/users/forrest
>
>
> On Sun, Aug 31, 2014 at 7:10 AM, Bhattacharya,Souvik <[log in to unmask]> 
> wrote:
>
>> Hi Guys,
>>
>> I have a time series dataframe which looks like
>>
>> 2014-02-05 2014-02-06 2014-02-07 2014-02-12 2014-02-14 2014-02-17
>> 2014-02-18 2014-02-19 ......
>>     0.0379    -0.0008     0.0352     0.0379     0.0392     0.0173
>>  0.0360     0.0371
>>
>> I want to compute moving standard deviation for every 5th day data 
>> from
>> this list. What I mean is that, I wish to select a sample in the 
>> form such
>> that sample1[1] = 2014-02-05  0.0379 , sample1[2] =2014-02-12
>> 0.0379.....and then find the std dev of this sample and then use a 
>> rolling
>> standard deviation to move on to the next date i.e. sample2[1] 
>> =2014-02-06
>> -0.0008 , sample2[2] =2014-02-12 0.0379 and find the standard 
>> deviation of
>> this list and so on. Since day available is irregular, I cannot use
>> seq(1:l, by = ). In rollapply, the function would take every 
>> consecutive
>> numbers to compute the standard deviation. Is there a way to sample 
>> every
>> 5th day data from this list in an efficient way, or modify the 
>> standard
>> deviation function somehow, to make it select every 5th day data and 
>> then
>> compute the standard deviation on the available data. Any suggestion 
>> in
>> this regard will be highly appreciated.
>>
>> Thanks
>> Souvik
>>
>> This list strives to be beginner friendly.  However, we still ask 
>> that you
>> PLEASE do read the posting guide http://www.R-project.org/
>> posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
> This list strives to be beginner friendly.  However, we still ask 
> that you
> PLEASE do read the posting guide 
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.

This list strives to be beginner friendly.  However, we still ask that you
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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