将逻辑值与以pandas/numpy的NaN进行比较

我想对两个熊猫系列的布尔值进行按元素或运算.还包括np.nan.

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

我尝试了三种方法，并且意识到根据方法，可以将表达式"np.nan或False"评估为True，False和np.nan.

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.

这些是我的示例系列:

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

方法1

使用pandas的|运算符:

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

方法2

使用numpy中的logical_or函数:

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

方法3

我定义了logical_or的矢量化版本，应该对数组进行逐行评估:

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)

我在两个系列上使用vectorized_or并将其输出(它是一个numpy数组)转换为pandas系列:

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

问题

我想知道这些结果的原因.
此答案解释了np.logical_or并说np.logical_or(np.nan, False)是True，但是为什么这仅在矢量化时才有效，而在矢量化时却不起作用在方法2中?以及如何解释方法1的结果?

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?