实现具有多个索引的字典的数据结构？

Question

我正在寻找一个数据结构,在两个不同的索引下保存相同的值,我可以通过任何一个访问数据.

例:

x = mysticalDataStructure()
x.add(1,'karl', dog)
x.add(2,'lisa', cat)

$ x[1].age
2
$ x['karl'].age
2
$ x[1].age = 4
$ x['karl'].age
4

是否有任何预先滚动,或者什么是最好的方法来滚动我自己(我需要通过索引访问(数字从0到n以1为增量),并通过字符串).

collections.ordereddict似乎没有通过位置快速随机访问,据我所知,我只能用迭代器走它,直到我到达元素i(我可以按正确的顺序插入).

Answer 1

有没有特别的原因你不能只使用字典:

x = {}
x[1] = x['karl'] = dog
x[2] = x['lisa'] = cat

然后你可以通过它们访问它.

如果你真的不想重复自己,你可以这样做:

class MysticalDataStructure(dict):
    def add(self, key1, key2, value):
        return self[key1] = self[key2] = value

x = MysticalDataStructure()
x.add(1, 'karl', dog)
x.add(2, 'lisa', cat)

Answer 2

class MultiKeyDict(object):

    def __init__(self, **kwargs):
        self._keys = {}
        self._data = {}
        for k, v in kwargs.iteritems():
            self[k] = v

    def __getitem__(self, key):
        try:
            return self._data[key]
        except KeyError:
            return self._data[self._keys[key]]

    def __setitem__(self, key, val):
        try:
            self._data[self._keys[key]] = val
        except KeyError:
            if isinstance(key, tuple):
               if not key:
                  raise ValueError(u'Empty tuple cannot be used as a key')
               key, other_keys = key[0], key[1:]
            else:
               other_keys = []
            self._data[key] = val
            for k in other_keys:
                self._keys[k] = key

    def add_keys(self, to_key, new_keys):
        if to_key not in self._data:
            to_key = self._keys[to_key]
        for key in new_keys:
            self._keys[key] = to_key


    @classmethod
    def from_dict(cls, dic):
        result = cls()
        for key, val in dic.items():
            result[key] = val
        return result

用法:

>>> d = MultiKeyDict(a=1, b=2)
>>> d['c', 'd'] = 3 # two keys for one value
>>> print d['c'], d['d']
3 3
>>> d['c'] = 4
>>> print d['d']
4
>>> d.add_keys('d', ('e',))
>>> d['e']
4
>>> d2 = MultiKeyDict.from_dict({ ('a', 'b'): 1 })
>>> d2['a'] = 2
>>> d2['b']
2