Comparing logical values to NaN in pandas/numpy
I want to do an element-wise OR operation on two pandas Series of boolean values. np.nan
s 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?
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.
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