Is there an efficient way to calculate rolling deviation from mean?

Question

I'd like to calculate the mean deviation '1/n[sum(Xi-mu)]' with rolling window. The 'mu' is rolling average. And Xi is rolling observation, too. Here is my sample code with window size n=10:

library(TTR)
dt<-rnorm(10000)
avg<-runMean(dt,n=10,cumulative=F)
df<-data.frame(dt,avg)

ls<-lapply(10:nrow(df),function(.){
    dev<-(df[(.-10+1):.,'dt']-df[.,'avg'])
    sk=mean(dev)
})

(p<-unlist(ls))

It seems the lapply is not an efficient way. Not sure what is alternative solution. Thank you anyone for any suggestion.

Take the mean of `abs(Xi-mu)`. – IRTFM Nov 12 '14 at 23:13 — IRTFM, Nov 12 '14 at 23:13

DatamineR · Answer 1 · 2014-11-12T23:40:51.600

2

Do you mean something like this (although the differences cancel out as simply explained here)?

library(zoo)
dt <- rnorm(1000)
rollapply(dt, 10, function(x) mean(x - mean(x)))

edited Nov 12 '14 at 23:40

answered Nov 12 '14 at 21:36

DatamineR

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It is not the variance. It the mean(Xi-mu). Variance is the mean((Xi-mu)^2) – YYY Nov 12 '14 at 22:02
2

But it doesn't make any sense, the deviations will cancel out and result in 0 – DatamineR Nov 12 '14 at 22:21
You can prove it modifying the function to: `function(x) mean(x - mean(x))` – DatamineR Nov 12 '14 at 22:29
Because it will be not be zero, but it is close. It will be used to calculate skewness. – YYY Nov 12 '14 at 22:35
it looks good. 50% faster than mine. Hopefully there is sth much more faster. – YYY Nov 12 '14 at 23:11
Variance is sum of squared deviations so all terms are non-zero and they" don't cancel out." If dealing with a MAD analysis, you use `abs`, – IRTFM Nov 12 '14 at 23:14
If they are not squared they do – DatamineR Nov 13 '14 at 00:12

Is there an efficient way to calculate rolling deviation from mean?

1 Answers1