Abiel Reinhart
2008-Jul-23 23:02 UTC
[R] Aggregating zoo object with NAs in multiple column
I would like to run an aggregation on a zoo object that has multiple series in it, with one of more series having NA values. The problem is that by default the aggregate function will produce an NA value in each aggregated period that contains an NA. For instance, if I run aggregate(x, as.yearmon(index(x)), mean) on the example object "x" which is printed below, I will just get a bunch of NAs for January. This behavior is perfectly logical. The problem is that if I try to use the na.omit() function, it will throw away the entire line if even one series has an NA value. For example, in the table below, you can see that running na.omit() will throw out periods 2001-01-06 through 2001-01-10. But since each of these lines contain many non-NA readings, we are throwing away real information that should be used in the calculation of the means for January. The mean for column B should include non-NA value for the month, but since A has a NA value on January 6, the January 6 value for B will be dropped as well. Same thing for columns C, D, and E. I suppose one solution would be to break the object into five one-series objects, run aggregate(na.omit(item), as.yearmon(index(na.omit(item))), mean) on each of them, then bind them back together, but this is rather annoying. Is there a better way? Thanks. Abiel a b c d e 2001-01-01 0.5183099 0.62792449 0.90859932 0.56578026 0.3991120 2001-01-02 0.2759420 0.96788392 0.30789409 0.76159986 0.3122280 2001-01-03 0.3263367 0.41224859 0.69756281 0.27406235 0.6902459 2001-01-04 0.3681782 0.41167564 0.02734471 0.39348676 0.8370692 2001-01-05 0.2550825 0.65790206 0.65134885 0.92537263 0.4143775 2001-01-06 NA 0.09076128 0.35209944 0.70821994 0.6659275 2001-01-07 0.4749008 NA 0.73579892 0.67311239 0.2155689 2001-01-08 0.7314498 0.56542607 NA 0.37529408 0.9313593 2001-01-09 0.5560702 0.47944318 0.01946189 NA 0.7055763 2001-01-10 0.4848510 0.12003527 0.31297935 0.41487588 NA 2001-01-11 0.0902985 0.88107285 0.33374604 0.26173483 0.3062338 2001-01-12 0.3664127 0.35366508 0.97760256 0.90784835 0.7399498 2001-01-13 0.6394206 0.05157520 0.38823937 0.92289256 0.6464278 2001-01-14 0.1949957 0.29738760 0.25224214 0.00024017 0.1228440 2001-01-15 0.7723980 0.99391775 0.22869908 0.97916413 0.1066641 [[alternative HTML version deleted]]
Achim Zeileis
2008-Jul-24 00:26 UTC
[R] Aggregating zoo object with NAs in multiple column
On Wed, 23 Jul 2008, Abiel Reinhart wrote:> I would like to run an aggregation on a zoo object that has multiple series > in it, with one of more series having NA values. The problem is that by > default the aggregate function will produce an NA value in each aggregated > period that contains an NA. For instance, if I run aggregate(x, > as.yearmon(index(x)), mean) on the example object "x" which is printed > below, I will just get a bunch of NAs for January.This is not specific to zoo series, the function mean() always behaves like this. If you want to remove the NAs before, you have to pass the argument na.rm = TRUE to mean. The easiest way to do this is aggregate(x, as.yearmon, mean, na.rm = TRUE) Z> This behavior is perfectly logical. The problem is that if I try to use the > na.omit() function, it will throw away the entire line if even one series > has an NA value. For example, in the table below, you can see that running > na.omit() will throw out periods 2001-01-06 through 2001-01-10. But since > each of these lines contain many non-NA readings, we are throwing away real > information that should be used in the calculation of the means for January. > The mean for column B should include non-NA value for the month, but since A > has a NA value on January 6, the January 6 value for B will be dropped as > well. Same thing for columns C, D, and E. > > I suppose one solution would be to break the object into five one-series > objects, run aggregate(na.omit(item), as.yearmon(index(na.omit(item))), > mean) on each of them, then bind them back together, but this is rather > annoying. Is there a better way? > > Thanks. > > Abiel > > a b c d e > 2001-01-01 0.5183099 0.62792449 0.90859932 0.56578026 0.3991120 > 2001-01-02 0.2759420 0.96788392 0.30789409 0.76159986 0.3122280 > 2001-01-03 0.3263367 0.41224859 0.69756281 0.27406235 0.6902459 > 2001-01-04 0.3681782 0.41167564 0.02734471 0.39348676 0.8370692 > 2001-01-05 0.2550825 0.65790206 0.65134885 0.92537263 0.4143775 > 2001-01-06 NA 0.09076128 0.35209944 0.70821994 0.6659275 > 2001-01-07 0.4749008 NA 0.73579892 0.67311239 0.2155689 > 2001-01-08 0.7314498 0.56542607 NA 0.37529408 0.9313593 > 2001-01-09 0.5560702 0.47944318 0.01946189 NA 0.7055763 > 2001-01-10 0.4848510 0.12003527 0.31297935 0.41487588 NA > 2001-01-11 0.0902985 0.88107285 0.33374604 0.26173483 0.3062338 > 2001-01-12 0.3664127 0.35366508 0.97760256 0.90784835 0.7399498 > 2001-01-13 0.6394206 0.05157520 0.38823937 0.92289256 0.6464278 > 2001-01-14 0.1949957 0.29738760 0.25224214 0.00024017 0.1228440 > 2001-01-15 0.7723980 0.99391775 0.22869908 0.97916413 0.1066641 > > [[alternative HTML version deleted]] > > ______________________________________________ > R-help at r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > >