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Using rollmean when there are missing values (NA)

Tags:

r

zoo

xts

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
like image 576
Alex Avatar asked Jul 20 '13 17:07

Alex


2 Answers

From ?rollmean

The default method of ‘rollmean’ does not handle inputs that contain ‘NA’s. In such cases, use ‘rollapply’ instead.

like image 150
GSee Avatar answered Nov 12 '22 03:11

GSee


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'.

like image 41
JKim Avatar answered Nov 12 '22 01:11

JKim