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