一只猴子如何在python中修补一个函数? [英] How does one monkey patch a function in python?

问题描述

我无法用另一个函数替换来自不同模块的函数，这让我发疯.

I'm having trouble replacing a function from a different module with another function and it's driving me crazy.

假设我有一个看起来像这样的模块 bar.py:

Let's say I have a module bar.py that looks like this:

from a_package.baz import do_something_expensive

def a_function():
    print do_something_expensive()

而且我还有一个看起来像这样的模块:

And I have another module that looks like this:

from bar import a_function
a_function()

from a_package.baz import do_something_expensive
do_something_expensive = lambda: 'Something really cheap.'
a_function()

import a_package.baz
a_package.baz.do_something_expensive = lambda: 'Something really cheap.'
a_function()

我希望得到结果:

Something expensive!
Something really cheap.
Something really cheap.

但我得到了这个:

Something expensive!
Something expensive!
Something expensive!

我做错了什么?

推荐答案

思考 Python 命名空间的工作原理可能会有所帮助:它们本质上是字典.所以当你这样做时:

It may help to think of how Python namespaces work: they're essentially dictionaries. So when you do this:

from a_package.baz import do_something_expensive
do_something_expensive = lambda: 'Something really cheap.'

这样想:

do_something_expensive = a_package.baz['do_something_expensive']
do_something_expensive = lambda: 'Something really cheap.'

希望你能意识到为什么这不起作用 :-) 一旦你将一个名字导入到命名空间中，你导入的命名空间中的名字的值 from 是无关紧要的.您只是在上面的本地模块的命名空间或 a_package.baz 的命名空间中修改 do_something_expensive 的值.但是因为 bar 直接导入 do_something_expensive ，而不是从模块命名空间引用它，你需要写入它的命名空间:

Hopefully you can realize why this doesn't work then :-) Once you import a name into a namespace, the value of the name in the namespace you imported from is irrelevant. You're only modifying the value of do_something_expensive in the local module's namespace, or in a_package.baz's namespace, above. But because bar imports do_something_expensive directly, rather than referencing it from the module namespace, you need to write to its namespace:

import bar
bar.do_something_expensive = lambda: 'Something really cheap.'

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