I've encountered something weird about numpy.void.
arr = np.empty ((len(results),), dtype=[('deltaf', float),
('quantize', [('int', int), ('frac',
int)])])
for i,r in enumerate (results):
arr[i] = (r[0]['deltaf'],
tuple(r[0]['quantize_mf']))
from collections import defaultdict, namedtuple
experiments = defaultdict(list)
testcase = namedtuple ('testcase', ['quantize'])
for e in arr:
experiments[testcase(e['quantize'])].append (e)
Now it seems that when e['quantize'] is used as a dictionary key, equal values
are not compared as equal:
In [36]: experiments
Out[36]: defaultdict(<type 'list'>, {testcase(quantize=(0, 0)): [(1.25, (0,
0))], testcase(quantize=(0, 0)): [(1.25, (0, 0))], testcase(quantize=(0, 0)):
[(1.25, (0, 0))]})
See, there are 3 'testcases' inserted, all with keys quantize=(0,0).
In [37]: e['quantize']
Out[37]: (0, 0)
In [38]: type(e['quantize'])
Out[38]: <type 'numpy.void'>
There's something weird here. If instead I do:
for e in arr:
experiments[testcase(tuple(e['quantize']))].append (e)
that is, convert e['quantize'] to a tuple before using it as a key, I get the
expected behavior:
In [40]: experiments
Out[40]: defaultdict(<type 'list'>, {testcase(quantize=(0, 0)): [(1.25, (0,
0)),
(1.25, (0, 0)), (1.25, (0, 0))]})
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