[R] NAs and row/column calculations

[Jim Bouldin]

> 
> On 12/03/2010, at 11:25 AM, Jim Bouldin wrote:
> 
> > 
> > I continue to have great frustrations with NA values--in particular
> making
> > summary calculations on rows or cols of a matrix containing them.  For
> > example, why does:
> > 
> >> a = matrix(1:30,nrow=5)
> >> is.na(a[c(1:2),c(3:4)]);a
> >     [,1] [,2] [,3] [,4] [,5] [,6]
> > [1,]    1    6   NA   NA   21   26
> > [2,]    2    7   NA   NA   22   27
> > [3,]    3    8   13   18   23   28
> > [4,]    4    9   14   19   24   29
> > [5,]    5   10   15   20   25   30
> >> apply(a[!is.na(a)],2,sum)
> > 
> > give me this:
> > 
> > "Error in apply(a[!is.na(a)], 2, sum) : dim(X) must have a positive
> length"
> > 
> > when
> > 
> >> dim(a)
> > [1] 5 6
> > 
> > What is the trick to calculating summary values from rows or columns
> > containing NAs?  Drives me nuts.  More nuts that is.
> 
> When you do a[!is.na(a)] you get a ***vector*** --- not a matrix.
> ``Obviously''!!!  

Well, obvious to you maybe, or someone who's done it before, but not to me.

The non-missing values of a cannot be arranged in
> a 5 x 6 matrix; there are only 26 of them.  So (as my late Uncle
> Stanley would have said) ``What the hell do you expect?''.

Silly me, I expected, based on (1) previous experience doing summary calcs
on subsets of a matrix using exactly that style of command, and (2) the
fact that dim(a) returns: [1] 5 6, and (3) the fact that a help search
under the "apply" function gives NO INDICATION of any possible use of the
na.rm command, AND (4) a help search on "na.action" does not even mention
na.rm, that:
 
> apply(a[!is.na(a)],2,sum)

would sum the non-NA elements of matrix a, by columns.  Terribly faulty
reasoning on my part, obviously.


> 
> The ``trick'' is to remove the NAs at the summing stage:
> 
> apply(a,2,sum,na.rm=TRUE)
> 
> Not all that tricky.
> 
> 	cheers,
> 
> 		Rolf Turner
> 
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Jim Bouldin, PhD
Research Ecologist
Department of Plant Sciences, UC Davis
Davis CA, 95616
530-554-1740