Question

我有以下意外行为

import numpy as np
class Test:
    def __radd__(self, other):
        print(f'value: {other}')

[1,2,3] + Test()
# prints: value: [1,2,3]
np.array([1,2,3]) + Test()
# prints
# value: 1
# value: 2
# value: 3

我希望第二次添加的行为方式与第一次相同，但事实并非如此。我能看到的唯一逻辑解释是，numpy +运算符首先以某种方式迭代参数，并尝试将它们中的每一个添加到Test()，第二个加法(int + Test)回退到Test.__radd__

所以

还有其他任何我缺失的解释吗？
如果没有，a + b如何在numpy中工作
是否相当于np.add(a, b)？

Answer 1

这是因为NumPy 'broadcasting'。

您的解释非常正确，因为np.add可以看到in the docs。

备注

就阵列广播而言，相当于x1 + x2。

如果您使用NumPy并查看np.array与内置列表的不同之处，我发现这更有意义。

Python 3.5.2 (default, Jul 6 2016, 16:37:16) Type 'copyright', 'credits' or 'license' for more information IPython 6.2.1 -- An enhanced Interactive Python. Type '?' for help. In [1]: import numpy as np In [2]: [1, 2, 3] + [1, 2, 3] Out[2]: [1, 2, 3, 1, 2, 3] In [3]: np.array([1, 2, 3]) + np.array([1, 2, 3]) Out[3]: array([2, 4, 6])

查看NDArrayOperatorsMixin implements the special methods for almost all of Python's builtin operators defined in the `operator` module的源代码here。看起来__add__, __radd__, __iadd__都设置为.add umath函数。但是我不确定实际的ndarray是否使用了混合，我认为你必须深入研究C代码以弄清楚它是如何处理的。

Answer 2

If I expand your class a bit we may get more of an idea of when __add__ is used, and when __radd__:

class Test:
    def __radd__(self, other):
        print(f'r value: {other}, {type(other)}')
        return f'{other}'
    def __add__(self, other):
        print(f'a value: {other}, {type(other)}')
        return other+3

With a list

In [285]: [1,2,3]+Test()
r value: [1, 2, 3], <class 'list'>    # use Test.__radd__
Out[285]: '[1, 2, 3]'
In [286]: Test()+[1,2,3]             # tries to use Test.__add__
a value: [1, 2, 3], <class 'list'>
....
<ipython-input-280-cd3f564be47a> in __add__(self, other)
      5     def __add__(self, other):
      6         print(f'a value: {other}, {type(other)}')
----> 7         return other+3
      8 
TypeError: can only concatenate list (not "int") to list

With an array:

In [287]: np.arange(3)+Test()     # use Test.__radd__ with each array element
r value: 0, <class 'int'>
r value: 1, <class 'int'>
r value: 2, <class 'int'>
Out[287]: array(['0', '1', '2'], dtype=object)
In [288]: Test()+np.arange(3)
a value: [0 1 2], <class 'numpy.ndarray'>
Out[288]: array([3, 4, 5])    # use Test.__add__ on whole array

With itself, a double use of Test.__add__:

In [289]: Test()+Test()
a value: <__main__.Test object at 0x7fc33a5a7a20>, <class '__main__.Test'>
a value: 3, <class 'int'>
Out[289]: 6

As I commented it can be tricky sorting out the __add__ v __radd__ delegation, and separating that from ndarray action.

add with Test() second gives the same output as [287]:

In [295]: np.add(np.arange(3),Test())
r value: 0, <class 'int'>
r value: 1, <class 'int'>
r value: 2, <class 'int'>
Out[295]: array(['0', '1', '2'], dtype=object)

np.add with Test() first is different from [288] above:

In [296]: np.add(Test(),np.arange(3))
a value: 0, <class 'int'>
a value: 1, <class 'int'>
a value: 2, <class 'int'>
Out[296]: array([3, 4, 5], dtype=object)

Answer 3

你的直觉是正确的。 Numpy添加了每个元素。您也可以使用基元来查看此行为：

np.array([1,2,3]) + 1  # [2,3,4]

numpy添加如何工作？

3 个答案: