I have a data set which has a couple of NA
in it. I take a rolling mean and expect that when there is no NA
in the window, the rolling mean should produce a number as opposed to NA
, however, rollmeanr
in zoo
does not seem to do this. Example:
require(zoo)
z = zoo(cbind(a=0:10, b=c(NA,10:1), c=sample(1:11,11)), 1:11)
rollmeanr(z, k=3, fill=NA)
a b c
1 NA NA NA
2 NA NA NA
3 1 NA 3.333333
4 2 NA 4.666667
5 3 NA 4.000000
6 4 NA 6.333333
7 5 NA 7.000000
8 6 NA 9.333333
9 7 NA 8.333333
10 8 NA 8.666667
11 9 NA 5.666667
rollapply(z, width=3, FUN=mean, by=1, by.column=TRUE, fill=NA, align="right")
a b c
1 NA NA NA
2 NA NA NA
3 1 NA 3.333333
4 2 9 4.666667
5 3 8 4.000000
6 4 7 6.333333
7 5 6 7.000000
8 6 5 9.333333
9 7 4 8.333333
10 8 3 8.666667
11 9 2 5.666667
I would expect these two calls to generate the same result. Please comment. Some session info:
sessionInfo()
R version 3.0.1 (2013-05-16)
Platform: x86_64-unknown-linux-gnu (64-bit)
locale:
[1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C
[3] LC_TIME=en_US.UTF-8 LC_COLLATE=en_US.UTF-8
[5] LC_MONETARY=en_US.UTF-8 LC_MESSAGES=en_US.UTF-8
[7] LC_PAPER=C LC_NAME=C
[9] LC_ADDRESS=C LC_TELEPHONE=C
[11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] zoo_1.7-10
loaded via a namespace (and not attached):
[1] grid_3.0.1 lattice_0.20-15
From ?rollmean
The default method of ‘rollmean’ does not handle inputs that contain ‘NA’s. In such cases, use ‘rollapply’ instead.
Use 'partial=TRUE' option. The option makes it possible to calculate data with NA.
> rollapply(z, width=3, FUN=function(x) mean(x, na.rm=TRUE), by=1, by.column=TRUE, partial=TRUE, fill=NA, align="right")
a b c
1 0.0 NaN 1.000000
2 0.5 10.0 5.500000
3 1.0 9.5 4.333333
4 2.0 9.0 6.666667
5 3.0 8.0 4.666667
6 4.0 7.0 6.000000
7 5.0 6.0 7.000000
8 6.0 5.0 8.666667
9 7.0 4.0 8.333333
10 8.0 3.0 7.000000
11 9.0 2.0 5.000000
If you want to change 'NaN' in the first row to '0', modify 'fill=NA' to 'fill=0'.
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