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]]
>
>
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