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
wrote:
> Hello,
>
> I have a xts object, I would l
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,POSI
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