为什么python不使用__iadd__作为求和和链接运算符? [英] Why doesn't python take advantage of __iadd__ for sum and chained operators?

问题描述

我刚刚进行了一次有趣的测试:

I just conducted an interesting test:

~$ python3 # I also conducted this on python 2.7.6, with the same result
Python 3.4.0 (default, Apr 11 2014, 13:05:11) 
[GCC 4.8.2] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> class Foo(object):
...     def __add__(self, other):
...         global add_calls
...         add_calls += 1
...         return Foo()
...     def __iadd__(self, other):
...         return self
...
>>> add_calls = 0
>>> a = list(map(lambda x:Foo(), range(6)))
>>> a[0] + a[1] + a[2]
<__main__.Foo object at 0x7fb588e6c400>
>>> add_calls
2
>>> add_calls = 0
>>> sum(a, Foo())
<__main__.Foo object at 0x7fb588e6c4a8>
>>> add_calls
6

很明显，__iadd__方法比__add__方法更有效，不需要分配新的类.如果添加的对象非常复杂，这将创建不必要的新对象，从而可能在我的代码中造成巨大的瓶颈.

Obviously, the __iadd__ method is more efficient than the __add__ method, not requiring the allocation of a new class. If my objects being added were sufficiently complicated, this would create unnecessary new objects, potentially creating huge bottlenecks in my code.

我希望在a[0] + a[1] + a[2]中，第一个操作将调用__add__，第二个操作将调用新创建的对象上的__iadd__.

I would expect that, in an a[0] + a[1] + a[2], the first operation would call __add__, and the second operation would call __iadd__ on the newly created object.

为什么python无法对此进行优化?

Why doesn't python optimize this?

推荐答案

__add__方法可以自由返回其他类型的对象，而__iadd__如果使用就地语义，则应返回self.这里不需要它们返回相同类型的对象，因此sum()不应依赖__iadd__的特殊语义.

The __add__ method is free to return a different type of object, while __iadd__ should, if using in-place semantics, return self. They are not required to return the same type of object here, so sum() should not rely on the special semantics of __iadd__.

您可以使用 functools.reduce()函数来实现您自己想要的功能:

You can use the functools.reduce() function to implement your desired functionality yourself:

from functools import reduce

sum_with_inplace_semantics = reduce(Foo.__iadd__, a, Foo())

演示:

>>> from functools import reduce
>>> class Foo(object):
...     def __add__(self, other):
...         global add_calls
...         add_calls += 1
...         return Foo()
...     def __iadd__(self, other):
...         global iadd_calls
...         iadd_calls += 1
...         return self
... 
>>> a = [Foo() for _ in range(6)]
>>> result = Foo()
>>> add_calls = iadd_calls = 0
>>> reduce(Foo.__iadd__, a, result) is result
True
>>> add_calls, iadd_calls
(0, 6)

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