Hi Gabor,
I appreciate your suggestion. I'm wondering if there's any faster
implementation that one could achieve. The dataset I have is about
400K rows and I have many of them. Just wondering if you have any
thoughts. Thanks.
Sakda
On Tue, Sep 8, 2009 at 6:57 AM, R_help Help<rhelp...@gmail.com> wrote:
> ---------- Forwarded message ----------
> From: Gabor Grothendieck <ggrothendi...@gmail.com>
> Date: Sun, Aug 30, 2009 at 11:08 PM
> Subject: Re: [R-SIG-Finance] Aggregating irregular time series
> To: R_help Help <rhelp...@gmail.com>
> Cc: r-sig-fina...@stat.math.ethz.ch, r-help@r-project.org
>
>
> Try this for the first question:
>
> neighborapply <- function(z, width, FUN) {
> out <- z
> ix <- seq_along(z)
> jx <- findInterval(time(z) + width, time(z))
> out[] <- mapply(function(i, j) FUN(c(0, z[seq(i+1, length =
> j-i)])), ix, jx)
> out
> }
>
> # test - corrected :948 in last line
>
> library(zoo)
> library(chron)
>
> Lines <- "Time x
> 10:00:00.021 20
> 10:00:00.224 20
> 10:00:01.002 19
> 10:00:02.948 20"
>
> z <- read.zoo(textConnection(Lines), header = TRUE, FUN = times)
>
> neighborapply(z, times("00:00:02"), sum)
>
> # and here is an alternative neighborapply
> # using loops. The non-loop solution does
> # have the disadvantage that it does as
> # readily extend to other situations which
> # is why we add this second solution to
> # the first question.
>
> neighborapply <- function(z, width, FUN) {
> out <- z
> tt <- time(z)
> for(i in seq_along(tt)) {
> for(j in seq_along(tt)) {
> if (tt[j] - tt[i] > width) break
> }
> if (j == length(tt) && tt[j] - tt[i] <= width) j <- j+1
> out[i] <- FUN(c(0, z[seq(i+1, length = j-i-1)]))
> }
> out
> }
>
> The second question can be answered along the lines
> of the first by modifying neighborapply in the loop solution
> appropriately.
>
> On Sun, Aug 30, 2009 at 9:38 PM, R_help Help<rhelp...@gmail.com> wrote:
>> Hi,
>>
>> I have a couple of aggregation operations that I don't know how to
>> accomplish. Let me give an example. I have the following irregular
>> time series
>>
>> time x
>> 10:00:00.021 20
>> 10:00:00.224 20
>> 10:00:01.002 19
>> 10:00:02:948 20
>>
>> 1) For each entry time, I'd like to get sum of x for the next 2
>> seconds (excluding itself). Using the above example, the output should
>> be
>>
>> time sumx
>> 10:00:00.021 39
>> 10:00:00.224 19
>> 10:00:01.442 20
>> 10:00:02:948 0
>>
>> 2) For each i-th of x in the series, what's the first passage time to
>> x[i]-1. I.e. the output should be
>>
>> time firstPassgeTime
>> 10:00:00.021 0.981
>> 10:00:00.224 0.778
>> 10:00:01.442 NA
>> 10:00:02:948 NA
>>
>> Is there any shortcut function that allows me to do the above? Thank you.
>>
>> adschai
>>
>> _______________________________________________
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>> https://stat.ethz.ch/mailman/listinfo/r-sig-finance
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>
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