Am 06.05.2021 um 13:02 schrieb Allin Cottrell:
On Thu, 6 May 2021, Sven Schreiber wrote:
Am 06.05.2021 um 10:05 schrieb Sven Schreiber:
Right, but I'm wondering whether this isn't inefficient because the ok
values are overwritten with themselves. Or aren't they internally?
BTW, shouldn't t
On Thu, 6 May 2021, Sven Schreiber wrote:
Am 06.05.2021 um 10:05 schrieb Sven Schreiber:
Right, but I'm wondering whether this isn't inefficient because the ok
values are overwritten with themselves. Or aren't they internally?
BTW, shouldn't this work?
matrix m = {1, NA}
eval misszero(m[2
Am 06.05.2021 um 10:05 schrieb Sven Schreiber:
Right, but I'm wondering whether this isn't inefficient because the ok
values are overwritten with themselves. Or aren't they internally?
BTW, shouldn't this work?
matrix m = {1, NA}
eval misszero(m[2])
I'm getting an error "is matrix not ser
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 r
Am 06.05.2021 08:38 schrieb Sven Schreiber:
Am 06.05.2021 um 07:42 schrieb atecon:
Hi all,
I just would like to ask whether this behavior is actually expected:
matrix m = {1, NA; 2, 3}
print m
eval cum(m)
? print m
m (2 x 2)
1 nan
2 3
? eval cum(m)
1 nan
3 nan
Am 06.05.2021 um 07:42 schrieb atecon:
Hi all,
I just would like to ask whether this behavior is actually expected:
matrix m = {1, NA; 2, 3}
print m
eval cum(m)
? print m
m (2 x 2)
1 nan
2 3
? eval cum(m)
1 nan
3 nan
I expected the entry (2,2) to be 3 as this
