字典树trie与分词

刷LeetCode会遇到字典树这道题，但是还不知道有这么多的应用

文本识别相关词

其实就是匹配词表，找到包含的最长词，我在最后附一个样例代码

分词

读苏神【中文分词系列】 1. 基于AC自动机的快速分词
发现字典树还用于分词，与AC自动机一起，AC自动机主要是KMP算法，python的实现库为pyahocorasick
使用方式就是add + make，可以点这里简单看下ahocorasick使用
苏神应用AC自动机主要做的分词，最大匹配法、最大概率组合
这里贴个最大概率组合的代码，更多直接看原博客

def max_proba_cut(sentence):
    """动态规划的思想，保留到某个点的最大path"""
    paths = {
     0: ([], 0)}
    end = 0
    # end_index,(str,prob) py2 j[0]为分词结果
    for i, j in dic.iter(sentence):
        start, end = 1+i-len(j[0]), i+1
        # 以start为终点的path不存在
        if start not in paths:
            last = max([i for i in paths if i < start])
            # 添加以start为终点的结果集
            paths[start] = (paths[last][0]+[sentence[last:start]], paths[last][1]-10)
        proba = paths[start][1]+j[1]
        if end not in paths or proba > paths[end][1]:
            paths[end] = (paths[start][0]+[j[0]], proba)
    if end < len(sentence):
        return paths[end][0] + [sentence[end:]]
    else:
        return paths[end][0]

字典树匹配词表

class Node(object):

    def __init__(self, is_end=False):
        self.is_end = False
        self.node = {
     }


class Trie(object):

    def __init__(self):
        self.size = 0
        self.root = Node()

    def add(self, word):
        """
        字典树添加word
        :param word:
        :return:
        """
        cur = self.root  # node
        for w in word:
            if cur.node.get(w) is None:
                cur.node[w] = Node()
            cur = cur.node[w]

        if not cur.is_end:
            cur.is_end = True
            self.size += 1

    def __len__(self):
        return self.size

    def __contains__(self, word):
        cur = self.root
        for w in word:
            if cur.node.get(w) is None:
                return False
            cur = cur.node[w]
        return cur.is_end

    def search1(self, title):
        """
        从一个字符串查找匹配到的产品词
        :param title:
        :return:
        """
        res = []
        cur = self.root
        n = len(title)
        for i, w in enumerate(title):
            if w in cur.node:
                tmp = cur.node[w]
                j = i + 1
                while not tmp.is_end and j < n:
                    if title[j] in tmp.node:
                        tmp = tmp.node[title[j]]
                        j += 1
                    else:
                        break
                if tmp.is_end:
                    # 沙发床 椰子汁
                    if j < n and title[j] in tmp.node and tmp.node[title[j]].is_end:
                        res.append(title[i:j+1])
                    else:
                        res.append(title[i:j])

        return list(set(res)) if res else res
    def search2(self, title):
        """
        从一个字符串查找匹配到的最长产品词
        :param title:
        :return:
        """
        res = []
        cur = self.root
        n = len(title)
        for i, w in enumerate(title):
            if w in cur.node:
                tmp = cur.node[w]
                j = i + 1
                while j < n:
                    if title[j] in tmp.node:
                        tmp = tmp.node[title[j]]
                        j += 1
                    else:
                        break
                if tmp.is_end:
                    res.append(title[i:j])

        return list(set(res)) if res else res