Am Freitag, 29. April 2011, 00:15:31 schrieb Muhali:
> nan is not 0, that's for sure. But in the nan package, nans are simply
> neglected, so it behaves as 0 for addition and as 1 for multiplication
> (sort of).
Neglegting the NaNs would make more sense to me (treating them as NA), but
that would be something like:
b = [2 5 NaN 7];
b(isnan(b)) = []
b =
2 5 7
cumsum(b)
ans =
2 7 14
which is different. And maybe that is a reason - nancumsum would be ambigous
(treat the NaNs as 0 or treat them as nonexisting?).
nansum is not ambigous.
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