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python实现有序字典

程序员文章站 2022-03-26 15:11:58
对于一个能够保存键值插入顺序的字典,是如何实现的? 主要有两点: 一个双向链表,用来记录字典的键值的插入顺序 一个键和链表节点的映射,主要用来删除键的时候,找到键对应的节点 python代码实现 ......

对于一个能够保存键值插入顺序的字典,是如何实现的?

主要有两点:

  一个双向链表,用来记录字典的键值的插入顺序

  一个键和链表节点的映射,主要用来删除键的时候,找到键对应的节点

python代码实现

class link:
    __slots__ = 'prev', 'next', 'key'


class orderdict:
    def __init__(self):
        self.root = link()
        self.map = {}
        self._node_map = {}
        self.root.next = self.root
        self.root.prev = self.root

    def __setitem__(self, key, value):
        if key in self._node_map:
            self.map[key] = value
        else:
            root = self.root
            last = root.prev
            link = link()
            link.prev, link.next, link.key = last, root, key
            last.next = link
            root.prev = link
            self._node_map[key] = link
            self.map[key] = value

    def __getitem__(self, item):
        return self.map[item]

    def __delitem__(self, key):
        del self.map[key]
        link = self._node_map.pop(key)
        link_prev, link_next = link.prev, link.next
        link_prev.next, link_next.prev = link_next, link_prev
        link.prev, link.next = none, none

    def pop(self):
        """
        lifo
        :return:
        """
        if not self._node_map:
            raise keyerror('dict is empty')
        root = self.root
        link = root.prev
        link_prev = link.prev
        link_prev.next = root
        root.prev = link_prev
        link.prev, link.next = none, none
        self._node_map.pop(link.key)
        return self.map.pop(link.key)

    def __iter__(self):
        root = self.root
        curr = root.next
        while curr != root:
            yield curr.key
            curr = curr.next

    def values(self):
        root = self.root
        curr = root.next
        while curr != root:
            yield self.map[curr.key]
            curr = curr.next
    def __str__(self):
root = self.root
curr = root.next
out = []
while curr != root:
out.append((curr.key, self.map[curr.key]))
curr = curr.next
return str(out)


if __name__ == '__main__':
d = orderdict()
d['a'] = '1'
d['b'] = '2'
d['c'] = 'c'
print(d)