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Interpolate xts
2 messages · Rustamali Manesiya, 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]]
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