[R] Interpolate xts

[Joshua Ulrich]

Use na.approx:

set.seed(21)
x <- xts(rnorm(10), Sys.time()-10:1)
is.na(x) <- 2:4
is.na(x) <- 8:9
na.approx(x)
na.spline(x)

Best,
--
Joshua Ulrich  |  FOSS Trading: www.fosstrading.com



On Tue, Jan 11, 2011 at 12:08 AM, Rustamali Manesiya
<rmanesiya at gmail.com> wrote:
> Hello,
>
>       I have a xts object, I would like to fill the NA with linear
> interpolated data. Can anyone please help.
>
>> str(zz)
> An ‘xts’ object from 2010-11-24 15:59:29 to 2010-11-24 16:00:00 containing:
>  Data: num [1:23401, 1] 312 312 312 312 312 ...
>  Indexed by objects of class: [POSIXct,POSIXt] TZ:
>  xts Attributes:
> List of 2
>  $ src    : chr "datafeed"
>  $ updated: POSIXct[1:1], format: "2011-01-08 00:33:05"
>
>>zz
> 2010-11-24 15:59:29 315.0300
> 2010-11-24 15:59:30       NA
> 2010-11-24 15:59:31       NA
> 2010-11-24 15:59:32       NA
> 2010-11-24 15:59:33       NA
> 2010-11-24 16:00:00 314.7000
>
>
> Rusty
>
>        [[alternative HTML version deleted]]
>
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
>