R-beta: sum of squares and NAs
Bill Simpson
wsimpson at uwinnipeg.ca
Fri Jul 3 22:48:43 CEST 1998
This surprised me. Is it the way it is supposed to be?
> x<-c(1,2,3,4,5)
> y<-c(1.2,1.3,1.4,1.5,NA)
> x-y
[1] -0.2 0.7 1.6 2.5 NA
> (x-y)^2
[1] 0.04 0.49 2.56 6.25 NaN
>>>so NA^2 = NaN? Why not still NA?
> sum((x-y)^2)
[1] NaN
>>>yes that is reasonable
So if you ever have a data set with missing observations (NAs), you can't
do any nlm() least squares fitting? That doesn't seem right.
Thanks for any help.
Bill Simpson
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