Next time please provide sample data in a form we can easily read in (look at
?dput for example)
If i understand this right:
yourData<-read.table(header=T,text="
date days rate
1996_01_02 15 5.74590
1996_01_02 50 5.67332
1996_01_02 78 5.60888
1996_01_02 169 5.47376
1996_01_02 260 5.35267
1996_01_02 351 5.27619
1996_01_03 14 5.74740
1996_01_03 49 5.67226
1996_01_03 77 5.60371
1996_01_03 168 5.47058
1996_01_03 259 5.34662
1996_01_03 350 5.26630
")
results<-sapply(unique(yourData$date),function(thisDate){
subSet <- yourData[yourData$date==thisDate,]
appr<-approx(subSet$days,subSet$rate,xout=seq(0,360,
by=30))
rates<-appr$y
names(rates)<-appr$x
rates
})
colnames(results)<-unique(yourData$date)
This gives 13 results per date though, and it can't interpolate the first and
last value. If you need those values that are not in-between, try spline
instead of approx (you never specified how you wanted to interpolate).
On 17.01.2013, at 15:50, beanbandit wrote:
> hi guys
>
> I need to interpolate values for the zero coupon yield curve. Following data
> is given
>
>
>
> date days rate
>
> 1996 01 02 15 5.74590
> 1996 01 02 50 5.67332
> 1996 01 02 78 5.60888
> 1996 01 02 169 5.47376
> 1996 01 02 260 5.35267
> 1996 01 02 351 5.27619
>
> 1996 01 03 14 5.74740
> 1996 01 03 49 5.67226
> 1996 01 03 77 5.60371
> 1996 01 03 168 5.47058
> 1996 01 03 259 5.34662
> 1996 01 03 350 5.26630
>
> For every day i have to interpolate 10 values, for example for maturities of
> 30,60 or 90 days. I have interpolate data for a one year period, 10
> interpolation values a day, so that equals 3600 values.
>
> what's the easiest way to implement this in R?
>
> please hlep!
>
>
>
>
> --
> View this message in context:
> http://r.789695.n4.nabble.com/Help-with-interpolation-tp4655843.html
> Sent from the R help mailing list archive at Nabble.com.
>
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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