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''!!! 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?''.
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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