Comparing numpy arrays containing NaN

Question

For my unittest, I want to check if two arrays are identical. Reduced example:

a = np.array([1, 2, np.NaN]) b = np.array([1, 2, np.NaN])  if np.all(a==b):     print 'arrays are equal' 

This does not work because nan != nan. What is the best way to proceed?

Answer 1

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.

Answer 2

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.