Numpy's 'True' is not the same 'True' as Python's 'True' and therefor is
fails:
>>> import numpy as np
>>> a = np.array([True, True, False])
>>> a[:]
array([ True, True, False], dtype=bool)
>>> a[0]
True
>>> a[0]==True
True
>>> a[0] is True
False
>>> type(a[0])
<type 'numpy.bool_'>
>>> type(True)
<type 'bool'>
Also, specifically, PEP 8 says DONT use 'is' or '==' for Booleans:
Don't compare boolean values to True or False using ==:
Yes: if greeting:
No: if greeting == True:
Worse: if greeting is True:
An empty numpy array does test falsey just as an empty Python list or empty dict does:
>>> [bool(x) for x in [[],{},np.array([])]]
[False, False, False]
Unlike Python, a numpy array of a single falsey element does test falsey:
>>> [bool(x) for x in [[False],[0],{0:False},np.array([False]), np.array([0])]]
[True, True, True, False, False]
But you cannot use that logic with a numpy array with more than one element:
>>> bool(np.array([0,0]))
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()
So the 'spirit' of PEP 8 with Numpy is probably to only test each element's truthiness:
>>> np.where(np.array([0,0]))
(array([], dtype=int64),)
>>> np.where(np.array([0,1]))
(array([1]),)
Or use any
:
>>> np.array([0,0]).any()
False
>>> np.array([0,1]).any()
True
And be aware that this is not what you expect:
>>> bool(np.where(np.array([0,0])))
True
Since np.where
is returning a nonempty tuple.