Comparing logical values to NaN in pandas/numpy

Question

I want to do an element-wise OR operation on two pandas Series of boolean values. np.nans are also included.

I have tried three approaches and realized that the expression "np.nan or False" can be evaluted to True, False, and np.nan depending on the approach.

These are my example Series:

series_1 = pd.Series([True, False, np.nan])
series_2 = pd.Series([False, False, False])

Approach #1

Using the | operator of pandas:

In [5]: series_1 | series_2
Out[5]: 
0     True
1    False
2    False
dtype: bool

Approach #2

Using the logical_or function from numpy:

In [6]: np.logical_or(series_1, series_2)
Out[6]: 
0     True
1    False
2      NaN
dtype: object

Approach #3

I define a vectorized version of logical_or which is supposed to be evaluated row-by-row over the arrays:

@np.vectorize
def vectorized_or(a, b):
   return np.logical_or(a, b)

I use vectorized_or on the two series and convert its output (which is a numpy array) into a pandas Series:

In [8]:  pd.Series(vectorized_or(series_1, series_2))
Out[8]: 
0     True
1    False
2     True
dtype: bool

Question

I am wondering about the reasons for these results.
This answer explains np.logical_or and says np.logical_or(np.nan, False) is be True but why does this only works when vectorized and not in Approach #2? And how can the results of Approach #1 be explained?

Answer 1

first difference : | is np.bitwise_or. it explains the difference between #1 and #2.

Second difference : since serie_1.dtype if object (non homogeneous data), operations are done row by row in the two first cases.

When using vectorize ( #3):

The data type of the output of vectorized is determined by calling the function with the first element of the input. This can be avoided by specifying the otypes argument.

For vectorized operations, you quit the object mode. data are first converted according to first element (bool here, bool(nan) is True) and the operations are done after.