如何Python函数访问自己的属性？ [英] how can python function access its own attributes?

问题描述

是有可能访问在函数范围内的Python函数对象的属性？

例如。让我们

def f():
    return SOMETHING

f._x = "foo"
f()           # -> "foo"

现在，有些事情是，如果我们希望有_x属性内容为foo回来了？如果它甚至有可能（只是）

now, what SOMETHING has to be, if we want to have the _x attribute content "foo" returned? if it's even possible (simply)

感谢

更新：

我想下面的工作还：

g = f
del f
g()          # -> "foo"

更新2：

声明，这是不可能的（如果它是的情况），以及为什么，是比提供一种方法它如何假例如更令人满意用不同的对象不是函数

Statement that it is not possible (if it is the case), and why, is more satisfying than providing a way how to fake it e.g. with a different object than a function

推荐答案

请功能的默认参数之一是函数本身的引用。

Solution

Make one of the function's default arguments be a reference to the function itself.

def f(self):
    return self.x
f.func_defaults = (f,)

实例：

>>> f.x = 17
>>> b = f
>>> del f
>>> b()
17

说明

楼主希望不需要全局名称查找的溶液。简单的解决方案

Explanation

The original poster wanted a solution that does not require a global name lookup. The simple solution

def f():
    return f.x

执行全局变量 F的在每次调用，它不符合要求的查询。如果˚F被删除，则函数失败。更复杂的检查建议以同样的方式失败。

performs a lookup of the global variable f on each call, which does not meet the requirements. If f is deleted, then the function fails. The more complicated inspect proposal fails in the same way.

我们需要的是执行的早期绑定的并存储对象本身绑定的参考。以下是概念上我们在做什么：

What we want is to perform early binding and store the bound reference within the object itself. The following is conceptually what we are doing:

def f(self=f):
    return self.x

在上面，自是一个局部变量，因此没有执行全球查找。但是，我们不能写code原样，因为˚F尚未确定，当我们试图绑定的默认值自它。相反，我们设置后，默认值f 定义。

In the above, self is a local variable, so no global lookup is performed. However, we can't write the code as-is, because f is not yet defined when we try to bind the default value of self to it. Instead, we set the default value after f is defined.

下面是一个简单的装饰为你做到这一点。请注意，自参数必须最后一个出现，不像方法，其中自第一。这也意味着，你必须给一个默认值，如果你的任何其他参数取默认值。

Here's a simple decorator to do this for you. Note that the self argument must come last, unlike methods, where self comes first. This also means that you must give a default value if any of your other arguments take a default value.

def self_reference(f):
    f.func_defaults = f.func_defaults[:-1] + (f,)
    return f

@self_reference
def foo(verb, adverb='swiftly', self=None):
    return '%s %s %s' % (self.subject, verb, adverb)

例如：

>>> foo.subject = 'Fred'
>>> bar = foo
>>> del foo
>>> bar('runs')
'Fred runs swiftly'

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