comparing numpy arrays containing NaN

The easiest way is use numpy.allclose() method, which allow to specify the behaviour when having nan values. Then your example will look like the following:

a = np.array([1, 2, np.nan])
b = np.array([1, 2, np.nan])

if np.allclose(a, b, equal_nan=True):
    print('arrays are equal')

Then arrays are equal will be printed.

You can find here the related documentation

Alternatively you can use numpy.testing.assert_equal or numpy.testing.assert_array_equal with a try/except:

In : import numpy as np

In : def nan_equal(a,b):
...:     try:
...:         np.testing.assert_equal(a,b)
...:     except AssertionError:
...:         return False
...:     return True

In : a=np.array([1, 2, np.NaN])

In : b=np.array([1, 2, np.NaN])

In : nan_equal(a,b)
Out: True

In : a=np.array([1, 2, np.NaN])

In : b=np.array([3, 2, np.NaN])

In : nan_equal(a,b)
Out: False

Edit

Since you are using this for unittesting, bare assert (instead of wrapping it to get True/False) might be more natural.

For versions of numpy prior to 1.19, this is probably the best approach in situations that don't specifically involve unit tests:

>>> ((a == b) | (numpy.isnan(a) & numpy.isnan(b))).all()
True

However, modern versions provide the array_equal function with a new keyword argument, equal_nan, which fits the bill exactly.

This was first pointed out by flyingdutchman; see his answer below for details.