[Numpy-discussion] Re: Comparison between numpy scalars returns numpy bool class and not native python bool class

June 27, 2024


      On Thu, Jun 27, 2024 at 3:48 PM Aaron Meurer <asmeurer@gmail.com> wrote:
...
Apparently the reason this happens is that True, False, and None are
compared using 'is' in structural pattern matching (see
https://peps.python.org/pep-0634/#literal-patterns).
There's no way NumPy could avoid this. First off, Python won't even
let you subclass bool:
...
...
...
class mybool(bool):
...     pass
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: type 'bool' is not an acceptable base type
np.bool_ objects *do* compare equal to bool:
...
...
...
type(np.array(1) > 0)
<class 'numpy.bool_'>
(np.array(1) > 0) == True
True
but that doesn't matter because True is specifically special cased in
structural pattern matching.
The workaround is to use np.True_ and np.False_ in your pattern
...
...
...
match a > 1:
...     case np.True_ | np.False_:
...         print('python float')
...     case _:
...         print('Huh?: numpy float')
python float
Fortunately, since these compare equal to Python bool, this will also
work even if a > 1 is a normal True or False:
...
...
...
a = 1
import numpy as np
... a = np.float64(1)
... assert isinstance(a, float)
... match a > 1:
...     case np.True_ | np.False_:
...         print('python float')
...     case _:
...         print('Huh?: numpy float')
python float
Typo. This should have just been:
...
...
...
a = 1
match a > 1:
...     case np.True_ | np.False_:
...         print("matches python bool")
...     case _:
...         print("doesn't match")
matches python bool
Aaron Meurer
...
Aaron Meurer
On Thu, Jun 27, 2024 at 3:33 PM Stefano Miccoli via NumPy-Discussion
<numpy-discussion@python.org> wrote:
...
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_______________________________________________
NumPy-Discussion mailing list -- numpy-discussion@python.org
To unsubscribe send an email to numpy-discussion-leave@python.org
https://mail.python.org/mailman3/lists/numpy-discussion.python.org/
Member address: asmeurer@gmail.com