It is well known that ‘np.bool' is not interchangeable with python ‘bool’, and in fact 'issubclass(np.bool, bool)’ is false.
On the contrary, numpy floats are subclassing python
floats—'issubclass(np.float64, float) is true—so I’m wondering if the fact that
scalar comparison returns a np.bool breaks the Liskov substitution principle.
In fact ’(np.float64(1) > 0) is True’ is unexpectedly false.
I was hit by this behaviour because in python structural pattern matching, the
‘a > 1’ subject will not match neither ’True’ or ‘False’ if ‘a' is a numpy
scalar: In this short example
import numpy as np
a = np.float64(1)
assert isinstance(a, float)
match a > 1:
case True | False:
print('python float')
case _:
print('Huh?: numpy float’)
the default clause is matched. If we set instead ‘a = float(1)’, the first
clause will be matched. The surprise factor is quite high here, in my opinion.
(Let me add that ‘True', ‘False', ‘None' are special in python structural
pattern matching, because they are matched by identity and not by equality.)
I’m not sure if this behaviour can be avoided, or if we have to live with the
fact that numpy floats are to be kept well contained and never mixed with
python floats.
Stefano
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