为相同的字典值创建可交换元组键的最佳方法是什么?

2022-01-20 00:00:00 python dictionary tuples sorting

问题描述

def check():
    dict_choice_a = {(a, b) : value, (b, a) : value}  #(a, b) and (b, a) refer to the same value but repeted
    dict_choice_b = {tuple(sorted((a, b)) : value}  #not repetitive but unreadable
    dict_choice_a[(a, b)] = new_value #need to do twice to change value but more readable than dict_choice_b
    dict_choice_a[(b, a)] = new_value

    #value of both keys are always the same

我想创建一个 dictionary 元组键引用其值，键需要可交换为 (a, b) = (b, a)并且它们只引用相同的值.

I want to create a dictionary that has tuple keys referred to its values, that keys need to be exchangeable as (a, b) = (b, a) and they only refer to the same value.

这里的问题是:使 tulpe of keys 的元素可交换但也引用相同值的最佳方法是什么.

Here's the question is: what is the best way to make the element of tulpe of keys exchangeable but also refer to the same value.

此外，字符串也应该在解决方案中起作用.

Moreover, string should be also work in the solution.

解决方案

根据评论，您可以将 a 和 b 放入一个 frozenset，这是无序的:

Per the comments, you can put a and b into a frozenset, which is unordered:

dict_choice = {frozenset((a, b)): value}

<小时>

如果您需要它是自动的，您可以创建自己的 MutableMapping:

class MyDict(MutableMapping):

    def __init__(self, arg=None):
        self._map = {}
        if arg is not None:
            self.update(arg)

    def __getitem__(self, key):
        return self._map[frozenset(key)]

    def __setitem__(self, key, value):
        self._map[frozenset(key)] = value

    def __delitem__(self, key):
        del self._map[frozenset(key)]

    def __iter__(self):
        return iter(self._map)

    def __len__(self):
        return len(self._map)

使用中:

>>> d = MyDict([((1, 2), 'hello'), ((3, 4), 'world')])
>>> d[(2, 1)]
'hello'

但是请注意，这可能会对其他类型的键产生意外行为:

However note that this could have unexpected behaviour with other kinds of keys:

>>> d['hello'] = 'world'
>>> d['hole']
'world'
>>> d[1] = 2
Traceback (most recent call last):
  File "python", line 1, in <module>
  File "python", line 14, in __setitem__
TypeError: 'int' object is not iterable

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