[R-SIG-Finance] zoo: how to find for series x closest day in series y?
Gabor Grothendieck
ggrothendieck at gmail.com
Fri Apr 23 12:46:37 CEST 2010
Here is another solution where x and y are as in your post.
f <- function(u) which.min(abs(as.numeric(index(y)) - as.numeric(u)))
ix <- sapply(index(x), f)
cbind(x, y = coredata(y)[ix])
On Fri, Apr 23, 2010 at 5:50 AM, Gabor Grothendieck
<ggrothendieck at gmail.com> wrote:
> Example 4d on the sqldf home page shows how:
> http://sqldf.googlecode.com
>
> Here it is modified to work with your example data.
>
> (Note that its important to use different column names in DFx and DFy
> since it can otherwise foul up sqldf's class assignment heuristic in
> this case. If for some reason you must use names that are the same in
> both DFx and DFy then use the method = "raw" argument to sqldf and
> convert the date columns to Date yourself. See sqldf home page for
> more info.)
>
> library(zoo)
> library(sqldf)
>
> x <- zoo(1:3,as.Date(c("1992-12-13", "1997-05-12", "1997-07-13")))
> y <- zoo(1:5,as.Date(c("1992-12-15", "1992-12-16", "1997-05-10","1997-05-19",
> "1997-07-13")))
>
> DFx <- data.frame(x = coredata(x), xt = index(x))
> DFy <- data.frame(y = coredata(y), yt = index(y))
>
> out <- sqldf("select x, y, xt, yt from DFx x, DFy y
> where abs(xt - yt) =
> (select min(abs(xt - y2.yt)) from DFy y2)")
>
> ix <- !duplicated(out[,3])
> zoo(out[ix, 1:2], out[ix, 3])
>
>
> On Fri, Apr 23, 2010 at 5:10 AM, Matthieu Stigler
> <matthieu.stigler at gmail.com> wrote:
>> Hi
>>
>> I have two series, x and y, which don't have the same index. I then need to
>> find which in y is the closest day in x. How can do I do this in a clean
>> way? I have here a dirty for loop solution...
>> library(zoo)
>> #my series
>> x<-zoo(1:3,as.Date(c("1992-12-13", "1997-05-12", "1997-07-13")))
>> y<-zoo(1:5,as.Date(c("1992-12-15", "1992-12-16", "1997-05-10","1997-05-19",
>> "1997-07-13")))
>>
>> #but index is not always the same:
>> index(x)%in%index(y)
>>
>> #my (dirty?) solution
>> xnew<-x
>> for(i in which(!index(x)%in%index(y)))
>> xnew[i]<-which.min(abs(as.numeric(index(x)[i]-index(y))))
>> xnew
>>
>> Is there something cleaner one could use to find the closest available day?
>>
>> Thanks a lot!
>>
>> Matthieu
>>
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