I assume that, because NaN != NaN, even though both have the same hash value
(hash(NaN) == -32768), that Python treats any NaN double as a distinct key
in a dictionary.
In [76]: a = np.repeat(nan, 10)
In [77]: d = {}
In [78]: for i, v in enumerate(a):
....: d[v] = i
....:
....:
In [79]: d
Out[79]:
{nan: 0,
nan: 1,
nan: 6,
nan: 4,
nan: 3,
nan: 9,
nan: 7,
nan: 2,
nan: 8,
nan: 5}
I'm not sure if this ever worked in a past version of NumPy, however, I have
code which does a "group by value" and occasionally in the real world those
values are NaN. Any ideas or a way around this problem?
Thanks,
Wes
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