python - creating a boolean array which compares numpy elements to None

Question

I have a numpy array with dtype=object, and I want to create a boolean array identifying which elements are None. But it looks like None behaves differently...

a = np.array(['Duck','Duck','Duck','Goose',None,1,2,3,1,3,None,4])
print a == 'Duck'
print a == 3
print a == None

which results in

[ True  True  True False False False False False False False False False]
[False False False False False False False  True False  True False False]
False

Is there an "numpythonic" way to get a boolean array of the None elements? I can use

np.array([x is None for x in a])

but this seems like there should be a better way.

I definitely would not have expected ndarrays to behave differently for None... good catch! — SethMMorton
...especially because of this quote from the NumPy docs: "Each of the arithmetic operations (+, -, *, /, //, %, divmod(), ** or pow(), <<, >>, &, ^, |, ~) and the comparisons (==, <, >, <=, >=, !=) is equivalent to the corresponding universal function (or ufunc for short) in Numpy." I can only assume the np.equal is the ufunc for ==. — SethMMorton

Ashwini Chaudhary Ashwini Chaudhary · Accepted Answer · 2014-09-25T22:36:55

You can use numpy.equal:

In [20]: np.equal(a, None)
Out[20]: 
array([False, False, False, False,  True, False, False, False, False,
       False,  True, False], dtype=bool)

python - creating a boolean array which compares numpy elements to None

1 Answers