比较包含NaN的numpy数组 [英] comparing numpy arrays containing NaN

问题描述

对于我的单元测试，我想检查两个数组是否相同.简化示例:

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'

这不起作用，因为nan != nan. 最好的进行方法是什么?

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

推荐答案

或者，您也可以使用

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

修改

由于您正在使用它进行单元测试，因此裸露的assert(而不是将其包装成True/False)可能会更自然.

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

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