python实现有序字典
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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)