字典树原理详解及其Python实现

一、原理详解

1、初步介绍:
字典树又名前缀树,Trie树,是一种存储大量字符串的树形数据结构,经常被搜索引擎系统用于文本词频统计。
除此之外也常用于计算左右信息熵、计算点互信息。

下图演示了一个保存了8个单词的字典树的结构,8个单词分别是:“A”, “to”, “tea”, “ted”, “ten”, “i”, “in”, “inn”。

字典树原理详解及其Python实现_第1张图片
2、优势:
利用字符串的公共前缀来减少查询时间,最大限度地减少无谓的字符串比较,查询效率高。
相比于HashMap存储,在存储单词(和语种无关,任意语言都可以)的场景上,节省了大量的内存空间。

3、基本性质:
(1)、根节点不包含字符,除根节点外每一个节点都只包含一个字符;
(2)、从根节点到某一节点,路径上经过的字符连接起来,为该节点对应的字符串;
(3)、每个节点的所有子节点包含的字符都不相同。

二、python实现

1、创建字典树的节点

class TrieNode:
    def __init__(self):
        self.nodes = dict()  # 构建字典
        self.is_leaf = False

2、实现插入操作

def insert(self, word: str): 
        curr = self
        for char in word:
            if char not in curr.nodes:
                curr.nodes[char] = TrieNode()
            curr = curr.nodes[char]
        curr.is_leaf = True

3、实现查找操作

def search(self, word: str):
        curr = self
        for char in word:
            if char not in curr.nodes:
                return False
            curr = curr.nodes[char]
        return curr.is_leaf

4、完整代码:
完整代码参考自:https://blog.csdn.net/danengbinggan33/article/details/82151220

# -*- coding:utf-8 -*-
"""
Description:大变双向字典树
迭代次数默认最大999,可以增加但是没必要。其实能深到999层,那这个序列还是选择另外的处理方式吧。

@author: WangLeAi
@date: 2018/8/15
"""


class TrieNode(object):
    def __init__(self, value=None, count=0, parent=None):
        # 值
        self.value = value
        # 频数统计
        self.count = count
        # 父结点
        self.parent = parent
        # 子节点,{value:TrieNode}
        self.children = {}


class Trie(object):
    def __init__(self):
        # 创建空的根节点
        self.root = TrieNode()

    def insert(self, sequence):
        """
        基操,插入一个序列
        :param sequence: 列表
        :return:
        """
        cur_node = self.root
        for item in sequence:
            if item not in cur_node.children:
                # 插入结点
                child = TrieNode(value=item, count=1, parent=cur_node)
                cur_node.children[item] = child
                cur_node = child
            else:
                # 更新结点
                cur_node = cur_node.children[item]
                cur_node.count += 1

    def search(self, sequence):
        """
        基操,查询是否存在完整序列
        :param sequence: 列表
        :return:
        """
        cur_node = self.root
        mark = True
        for item in sequence:
            if item not in cur_node.children:
                mark = False
                break
            else:
                cur_node = cur_node.children[item]
        # 如果还有子结点,说明序列并非完整
        if cur_node.children:
            mark = False
        return mark

    def delete(self, sequence):
        """
        基操,删除序列,准确来说是减少计数
        :param sequence: 列表
        :return:
        """
        mark = False
        if self.search(sequence):
            mark = True
            cur_node = self.root
            for item in sequence:
                cur_node.children[item].count -= 1
                if cur_node.children[item].count == 0:
                    cur_node.children.pop(item)
                    break
                else:
                    cur_node = cur_node.children[item]
        return mark

    def search_part(self, sequence, prefix, suffix, start_node=None):
        """
        递归查找子序列,返回前缀和后缀结点
        此处简化操作,仅返回一位前后缀的内容与频数
        :param sequence: 列表
        :param prefix: 前缀字典,初始传入空字典
        :param suffix: 后缀字典,初始传入空字典
        :param start_node: 起始结点,用于子树的查询
        :return:
        """
        if start_node:
            cur_node = start_node
            prefix_node = start_node.parent
        else:
            cur_node = self.root
            prefix_node = self.root
        mark = True
        # 必须从第一个结点开始对比
        for i in range(len(sequence)):
            if i == 0:
                if sequence[i] != cur_node.value:
                    for child_node in cur_node.children.values():
                        self.search_part(sequence, prefix, suffix, child_node)
                    mark = False
                    break
            else:
                if sequence[i] not in cur_node.children:
                    for child_node in cur_node.children.values():
                        self.search_part(sequence, prefix, suffix, child_node)
                    mark = False
                    break
                else:
                    cur_node = cur_node.children[sequence[i]]
        if mark:
            if prefix_node.value:
                # 前缀数量取序列词中最后一词的频数
                if prefix_node.value in prefix:
                    prefix[prefix_node.value] += cur_node.count
                else:
                    prefix[prefix_node.value] = cur_node.count
            for suffix_node in cur_node.children.values():
                if suffix_node.value in suffix:
                    suffix[suffix_node.value] += suffix_node.count
                else:
                    suffix[suffix_node.value] = suffix_node.count
            # 即使找到一部分还需继续查找子结点
            for child_node in cur_node.children.values():
                self.search_part(sequence, prefix, suffix, child_node)


if __name__ == "__main__":
    trie = Trie()
    texts = [["葬爱", "少年", "葬爱", "少年", "慕周力", "哈哈"], ["葬爱", "少年", "阿西吧"], ["烈", "烈", "风", "中"], ["忘记", "了", "爱"],
             ["埋葬", "了", "爱"]]
    for text in texts:
        trie.insert(text)
    markx = trie.search(["忘记", "了", "爱"])
    print(markx)
    markx = trie.search(["忘记", "了"])
    print(markx)
    markx = trie.search(["忘记", "爱"])
    print(markx)
    markx = trie.delete(["葬爱", "少年", "王周力"])
    print(markx)
    prefixx = {}
    suffixx = {}
    trie.search_part(["葬爱", "少年"], prefixx, suffixx)
    print(prefixx)
    print(suffixx)

参考:
https://leetcode-cn.com/problems/short-encoding-of-words/solution/99-java-trie-tu-xie-gong-lue-bao-jiao-bao-hui-by-s/
https://blog.csdn.net/danengbinggan33/article/details/82151220

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