Am 06.05.2021 um 09:42 schrieb atecon:
Am 06.05.2021 08:38 schrieb Sven Schreiber:
I'd say it's expected because nans propagate. Basically you don't know
what 3 + nan is. To get what you want, I was going to say "do
cum(misszero(m))", but that doesn't work on a matrix. Maybe it should?
(But remember the old fact that a NaN and a missing value is not always
the same thing.)
True, 3 + nan remains unclear or is actually not defined. Fine for me if
that's how it is handled -- just wanted to know what's expected in this
case.
Replacing nan values in matrix can be easily done by:
<hansl>
m = ok(m) ? m : VALUE # VALUE is some numeric value
</hansl>
Right, but I'm wondering whether this isn't inefficient because the ok
values are overwritten with themselves. Or aren't they internally?
cheers
sven
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