Combine several bool pandas.Series with uncommon indices with an logical and - pandas

Question

The Problem:
I have a list of pandas.Series, where the series all have dates as index, but it is not guaranteed that they all have the same index.
The values are guaranteed to be bools (no NaNs possible).
The result i want to get is one pandas.Series where the index is the union of all indices found in the list of series. The value for each index should be the logical and of all series values, which contain the index.
Example:
A = pd.Series(index=[datetime(2015,05,01,20),
datetime(2015,05,01,20,15),
datetime(2015,05,01,20,30)],
data=[False, True, True])
B = pd.Series(index=[datetime(2015,05,01,20),
datetime(2015,05,01,20,30),
datetime(2015,05,01,20,45)],
data=[True, True, True])
series = [A, B]
A common index is datetime(2015,05,01,20) the result at this index should be False and True i.e. False.
An uncommon index is datetime(2015,05,01,20,45), it is only found in series B. The expected result is to be the value of B at this index, i.e. True.
The desired result in total looks like this:
result = pd.Series(index=[datetime(2015,05,01,20),
datetime(2015,05,01,20,15),
datetime(2015,05,01,20,30),
datetime(2015,05,01,20,45)],
data=[False, True, True, True])
My Approach:
I came up with a good start (I think) but cannot find the correct operation, it currently looks like this
result = None
for next in series:
if result is None:
result = next
else:
result = result.reindex(index=result.index | next.index)
# the next line sadly returns: ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().
result.loc[next.index] = result.loc[next.index] and next.loc[next.index] # sadly returns

Answer 1

I should have digged a little further before asking. I found a solution, which works for me and looks like the pandas-way of doing it, but I might stand corrected if an even more convenient way is presented!
result = None
for next in series:
if result is None:
result = next
else:
index = result.index | next.index
result = result.reindex(index, fill_value=True) & next.reindex(index, fill_value=True)

Answer 2

If I understand what you want, I'd concat the 2 series column-wise and then call a function row-wise that drops the NaN values and returns the logical and of the 2 columns or the lone column value:
In [231]:
df = pd.concat([A,B], axis=1)
def func(x):
l = x.dropna()
if len(l) > 1:
return l[0]&l[1]
return l.values[0]
df['result'] = df.apply(func, axis=1)
df
​
Out[231]:
0 1 result
2015-05-01 20:00:00 False True False
2015-05-01 20:15:00 True NaN True
2015-05-01 20:30:00 True True True
2015-05-01 20:45:00 NaN True True