如何在函数内部获取函数对象(Python) [英] How to get function object inside a function (Python)

问题描述

我想要类似的东西

def x():
  print get_def_name()

，但不一定知道x的名称.

but not necessarily know the name of x.

理想情况下，它将返回'x'，其中x是函数的名称.

Ideally it would return 'x' where x would be the name of the function.

推荐答案

Python中的函数是对象，并且碰巧这些对象确实具有包含其定义名称的属性:

Functions in Python are objects, and as it happens those objects do have an attribute containing the name they were defined with:

>>> def x():
...     pass
...
>>> print x.__name__
x

因此，一个简单的方法可能是这样:

So, a naïve approach might be this:

>>> def x():
...     print x.__name__
... 
>>> x()
x

那似乎行得通.但是，由于您必须知道函数内部的x名称才能执行此操作，因此您实际上并没有获得任何好处.您可能已经做到了:

That seems to work. However, since you had to know the name of x inside the function in order to do that, you haven't really gained anything; you might have well just have done this:

def x():
    print "x"

实际上，这比这更糟，因为__name__属性仅指的是函数定义的名称.如果它绑定到另一个名称，它将不会像您期望的那样起作用:

In fact, though, it's worse than that, because the __name__ attribute only refers to the name the function was defined with. If it gets bound to another name, it won't behave as you expect:

>>> y = x
>>> y()
x

更糟糕的是，如果原来的名称不再存在，它将根本无法使用:

Even worse, if the original name is no longer around, it won't work at all:

>>> del x
>>> y()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 2, in x
NameError: global name 'x' is not defined

第二个问题是您实际上可以解决的问题，尽管它并不漂亮.诀窍是编写一个装饰器，该装饰器可以将函数名称作为参数传递给它:

This second problem is one you can actually get around, although it's not pretty. The trick is to write a decorator that can pass the function's name into it as an argument:

>>> from functools import wraps
>>> def introspective(func):
...     __name__ = func.__name__
...     @wraps(func)
...     def wrapper(*args, **kwargs):
...         return func(__name__=__name__, *args, **kwargs)
...     return wrapper
... 
>>> @introspective
... def x(__name__):
...     print __name__
... 
>>> x()
x
>>> y = x
>>> y()
x
>>> del x
>>> y()
x

...尽管如您所见，您仍然只返回定义该函数的名称，而不是现在刚绑定到的函数名称.

... although as you can see, you're still only getting back the name the function was defined with, not the one it happens to be bound to right now.

在实践中，简短的(正确的)答案是不要这样做".这是Python的一个基本事实，即对象不知道它们的名称(或多个名称)绑定到-如果您认为函数需要该信息，则说明您做错了.

In practice, the short (and correct) answer is "don't do that". It's a fundamental fact of Python that objects don't know what name (or names) they're bound to - if you think your function needs that information, you're doing something wrong.

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