Question

我想比较考虑位置的两个numpy数组一元素。例如

[1, 2, 3]==[1, 2, 3]  -> True

[1, 2, 3]==[2, 1, 3]  -> False

我尝试了以下

    for index in range(list1.shape[0]):
        if list1[index] != list2[index]:
            return False
    return True

但我收到以下错误

ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()

但是，以下内容并非正确使用.any或.all

 numpy.any(numpy.array([1,2,3]), numpy.array([1,2,3]))


 numpy.all(numpy.array([1,2,3]), numpy.array([1,2,3]))

返回时

 TypeError: only length-1 arrays can be converted to Python scalars

我很困惑，有人可以解释我做错了吗

由于

Answer 1

您还可以使用array_equal：

In [11]: a = np.array([1, 2, 3])

In [12]: b = np.array([2, 1, 3])

In [13]: np.array_equal(a, a)
Out[13]: True

In [14]: np.array_equal(a, b)
Out[14]: False

这应该更高效，因为您不需要保留临时a==b ...

注意：对于要使用np.all而不是all的较大数组，请稍微考虑一下性能。 array_equal执行大约相同的，除非数组提前不同，然后它会更快，因为它可以提前失败：

In [21]: a = np.arange(100000)

In [22]: b = np.arange(100000)

In [23]: c = np.arange(1, 100000)

In [24]: %timeit np.array_equal(a, a)  # Note: I expected this to check is first, it doesn't
10000 loops, best of 3: 183 µs per loop

In [25]: %timeit np.array_equal(a, b)
10000 loops, best of 3: 189 µs per loop

In [26]: %timeit np.array_equal(a, c)
100000 loops, best of 3: 5.9 µs per loop

In [27]: %timeit np.all(a == b)
10000 loops, best of 3: 184 µs per loop

In [28]: %timeit np.all(a == c)
10000 loops, best of 3: 40.7 µs per loop

In [29]: %timeit all(a == b)
100 loops, best of 3: 3.69 ms per loop

In [30]: %timeit all(a == c) # ahem!
# TypeError: 'bool' object is not iterable

Answer 2

您可以将一组布尔值传递给all，例如：

>>> import numpy as np
>>> a = np.array([1, 2, 3])
>>> b = np.array([2, 1, 3])
>>> a == b
array([False, False,  True], dtype=bool)
>>> np.all(a==b) # also works with all for 1D arrays
False

请注意，对于小型数组，内置all比np.all快得多（np.array_equal仍然较慢）：

>>> timeit.timeit("all(a==b)", setup="import numpy as np; a = np.array([1, 2, 3]); b = np.array([2, 1, 3])")
0.8798369040014222
>>> timeit.timeit("np.all(a==b)", setup="import numpy as np; a = np.array([1, 2, 3]); b = np.array([2, 1, 3])")
9.980971871998918
>>> timeit.timeit("np.array_equal(a, b)", setup="import numpy as np; a = np.array([1, 2, 3]); b = np.array([2, 1, 3])")
13.838635700998566

但无法正确使用多维数组：

>>> a = np.arange(9).reshape(3, 3)
>>> b = a.copy()
>>> b[0, 0] = 42
>>> all(a==b)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()
>>> np.all(a==b)
False

对于较大的数组，np.all最快：

>>> timeit.timeit("np.all(a==b)", setup="import numpy as np; a = np.arange(1000); b = a.copy(); b[999] = 0")
13.581198551000853
>>> timeit.timeit("all(a==b)", setup="import numpy as np; a = np.arange(1000); b = a.copy(); b[999] = 0")
30.610838356002205
>>> timeit.timeit("np.array_equal(a, b)", setup="import numpy as np; a = np.arange(1000); b = a.copy(); b[999] = 0")
17.95089965599982

如何比较numpy数组元素考虑元素的位置？

2 个答案: