BUG: np.bool_(False) & pd.NA gives pd.NA

[jbrockmendel]

>>> False & pd.NA
False
>>> np.bool_(False) & pd.NA
<NA>

this surprised me; is it on purpose @jorisvandenbossche?

[jorisvandenbossche]

Yes, ideally those are consistent .. This looks like a bug. Similarly, using a numpy array (instead of numpy scalar) also gives this different result:

>>> pd.NA & False
False
>>> pd.NA & np.bool_(False)
<NA>
>>>  pd.NA & np.array([True, False])
array([<NA>, <NA>], dtype=object)

In the implementation of __and__, we are only considering the cases of other being True, False or NA, and so defer to numpy for numpy arrays or scalars (I am not entirely sure how we then end up with the above result, though? Because explicitly putting the NA in a numpy container like np.array([pd.NA], dtype=object) & np.array([True, False]) actually gives the correct result).

Looking at the other dunders, the arithmetic/comparison ops seems to be using helpers like _create_binary_propagating_op, which also handle numpy arrays for other. We should probably do something similar for the logical operators?

[jorisvandenbossche]

So what I don't understand is how we get this difference in behaviour:

>>> np.array(pd.NA, dtype=object) & np.array([True, False])
array([<NA>, False], dtype=object)

>>> pd.NA & np.array([True, False])
array([<NA>, <NA>], dtype=object)