[R-SIG-Finance] zoo: how to find for series x closest day in series y?
Matthieu Stigler
matthieu.stigler at gmail.com
Fri Apr 23 15:04:59 CEST 2010
Dear gabor
Thanks a lot for your valuable help!!! I finally used your second
solution, which is more say in "zoo spirit". As I actually needed not
the closest day, but the closest day AFTER, I changed it slightly to:
pos<-function(x) ifelse(x>=0, x, NA)
f2 <- function(u) which.min(pos(as.numeric(index(y)) - as.numeric(u)))
ix <- sapply(index(x), f2)
Thanks a lot again for your help!!
Mat
Gabor Grothendieck a écrit :
> 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
>>>
>>> _______________________________________________
>>> R-SIG-Finance at stat.math.ethz.ch mailing list
>>> https://stat.ethz.ch/mailman/listinfo/r-sig-finance
>>> -- Subscriber-posting only. If you want to post, subscribe first.
>>> -- Also note that this is not the r-help list where general R questions
>>> should go.
>>>
>>>
More information about the R-SIG-Finance
mailing list