在 Python 中检索对对象的引用列表 [英] Retrieving the list of references to an object in Python

问题描述

全部:

a = 1b = ac = b

现在我想获取一个标记了对象 1 的列表，即 [a, b, c].我怎么能这样做?

顺便说一句，如何在这里正式调用变量a"?到目前为止，我知道它是对象的对象标签"，但我不知道它的术语是什么.

谢谢！

为什么我需要这个:

a = b = c = 1打印 a, b, c1 1 1一 = 2打印 a, b, c2 1 1

在C等其他语言中，如果我重新分配a = 2，a，b，c应该是2，但是在python中，没有像引用这样的东西，所以改变abc的所有值的唯一方法是据我所知，a = b = c = 2，这就是为什么要获取对象的所有引用.

解决方案

你问的不是很实用，也不可能.这是一种疯狂的做法:

<预><代码>>>>一 = 1>>>b = a>>>c = b>>>当地人(){'a': 1, 'c': 1, 'b': 1, '__builtins__': <module '__builtin__' (builtin)>, '__package__': None, '__name__': '__main__', '__doc__': 无}>>>[key for key, value in locals().items() if value == 1]['a', 'c', 'b']>>>全局变量(){'a': 1, 'c': 1, 'b': 1, '__builtins__': <module '__builtin__' (builtin)>, '__package__': None, '__name__': '__main__', '__doc__': 无}>>>[key for key, value in globals().items() if value == 1]['a', 'c', 'b']

All:

a = 1
b = a
c = b

Now I want to get a list of object 1 tagged, which is [a, b, c]. How could I do this?

BTW, how to call variable "a" here officially? I know so far it is a "object tag" for the object, but I have no idea what is the term of it.

Thanks!

why do I need this:

a = b = c = 1 
print a, b, c 
1 1 1
a = 2
print a, b, c 
2 1 1

in other language such as C, a,b,c should be 2 if I re-assign a = 2, but in python, there's no such thing like reference, so the only way to change all the value of a b c is a = b = c = 2 so far as I know, that is why purposed to get all reference of an object.

解决方案

What you're asking isn't very practical and isn't possible. Here's one crazy way of doing it:

>>> a = 1
>>> b = a
>>> c = b
>>> locals()
{'a': 1, 'c': 1, 'b': 1, '__builtins__': <module '__builtin__' (built-in)>, '__package__': None, '__name__': '__main__', '__doc__': None}
>>> [key for key, value in locals().items() if value == 1]
['a', 'c', 'b']
>>> globals()
{'a': 1, 'c': 1, 'b': 1, '__builtins__': <module '__builtin__' (built-in)>, '__package__': None, '__name__': '__main__', '__doc__': None}
>>> [key for key, value in globals().items() if value == 1]
['a', 'c', 'b']

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