外部重写python模块函数

Question

是否可以在不修改给定模块的情况下覆盖内部模块功能？你知道吗

下面的例子是我能想象到的最好的简化。对于原来的问题，请看问题的结尾。你知道吗

我有一个具有以下结构的包：

a
├── b.py
├── c.py
└── __init__.py

其中b.py是

from c import c_func

def b_func():
    print('b.b_func')
    return c_func()

而c.py是

def c_func():
    print('c.c_func')
    return 'return_c'

我想修改外部主.py

import a
from a.b import b_func

print('Calling b_func without modification')
solution = b_func()
print(solution)

# Trying to modify the internal function
print('Calling b_func with modification')
old_c_func = a.c.c_func

def new_c_func(*args, **kwargs):
    print('do something in new_c_func')
    return('return_new_c')

a.c.c_func = new_c_func

solution = b_func()
print(solution)

前面的代码输出以下内容

Calling b_func without modification
b.b_func
c.c_func
return_c
Calling b_func with modification
b.b_func
c.c_func
return_c

但我希望

Calling b_func without modification
b.b_func
c.c_func
return_c
Calling b_func with modification
b.b_func
do something in new_c_func
return_new_c_func

最初的问题与Scipy的一个私有函数有关，但我假设我的问题的答案概括为以下问题：

import scipy

# Stack calls
# minimize calls scipy.optimize._minimize_trust_ncg
# See: https://github.com/scipy/scipy/blob/2526df72e5d4ca8bad6e2f4b3cbdfbc33e805865/scipy/optimize/_minimize.py#L463
# _minimize_trust_ncg calls scipy.optimize._trustregion._minimize_trust_region
# See: https://github.com/scipy/scipy/blob/2526df72e5d4ca8bad6e2f4b3cbdfbc33e805865/scipy/optimize/_trustregion_ncg.py#L39
from scipy.optimize import minimize

old_minimize_trust_region = scipy.optimize._trustregion._minimize_trust_region

def new_minimize_trust_region(*args, **kwargs):
    print('new function')
    return old_minimize_trust_region

scipy.optimize._trustregion._minimize_trust_region = new_minimize_trust_region

x0 = [2]
fun = lambda x: x**2 + 42
jac = lambda x: 2*x
hess = lambda x: 2

method = 'trust-ncg'
solution = minimize(fun, x0, method=method, jac=jac, hess=hess)

print(solution)

Answer 1

我通过更改要修改的函数的属性code找到了临时解决方案。以下代码按需要工作：

import a.b
import a.c

print('Calling b_func without modification')
solution = a.b.b_func()
print(solution)

# Trying to modify the internal function
print('Calling b_func with modification')

def new_c_func(*args, **kwargs):
    print('do something in new_c_func')
    return('return new_c_func')

a.c.c_func.__code__ = new_c_func.__code__

solution = a.b.b_func()
print(solution)

我仍然缺少一个完整的解释，也不知道代码是否合适，但它是按预期工作。我被Stackoverflow问题Can you patch just a nested function with closure, or must the whole outer function be repeated?启发了