在 python trie 中存储字数

Question

我拿了一个单词列表并将其放入 trie 中。我还想在里面存储字数以供进一步分析。最好的方法是什么？这是我认为将收集和存储频率的课程，但我不知道如何去做。你可以看到我的尝试，插入的最后一行是我尝试存储计数的地方。

class TrieNode:
    def __init__(self,k):
        self.v = 0
        self.k = k
        self.children = {}
    def all_words(self, prefix):
        if self.end:
            yield prefix
        for letter, child in self.children.items():
            yield from child.all_words(prefix + letter)
class Trie:
    def __init__(self):
        self.root = TrieNode()
    def __init__(self):
        self.root = TrieNode()
    
    def insert(self, word):
        curr = self.root
        for letter in word:
            node = curr.children.get(letter)
            if not node:
                node = TrieNode()
                curr.children[letter] = node
            curr.v += 1

    def insert_many(self, words):
        for word in words:
            self.insert(word)
    def all_words_beginning_with_prefix(self, prefix):
        cur = self.root
        for c in prefix:
            cur = cur.children.get(c)
            if cur is None:
                return  # No words with given prefix
        yield from cur.all_words(prefix)

我想存储计数，以便在我使用时

print(list(trie.all_words_beginning_with_prefix('prefix')))

我会得到这样的结果：

[(word, count), (word, count)]

Answer 1

插入时，在看到任何节点时，这意味着将在该路径中添加一个新词。因此增加该节点的 word_count。

class TrieNode:
    def __init__(self, char):
        self.char = char
        self.word_count = 0
        self.children = {}

    def all_words(self, prefix, path):
        if len(self.children) == 0:
            yield prefix + path
        for letter, child in self.children.items():
            yield from child.all_words(prefix, path + letter)


class Trie:
    def __init__(self):
        self.root = TrieNode('')

    def insert(self, word):
        curr = self.root
        for letter in word:
            node = curr.children.get(letter)
            if node is None:
                node = TrieNode(letter)
                curr.children[letter] = node
            curr.word_count += 1  # increment it everytime the node is seen at particular level.
            curr = node

    def insert_many(self, words):
        for word in words:
            self.insert(word)

    def all_words_beginning_with_prefix(self, prefix):
        cur = self.root
        for c in prefix:
            cur = cur.children.get(c)
            if cur is None:
                return  # No words with given prefix
        yield from cur.all_words(prefix, path="")

    def word_count(self, prefix):
        cur = self.root
        for c in prefix:
            cur = cur.children.get(c)
            if cur is None:
                return 0
        return cur.word_count


trie = Trie()
trie.insert_many(["hello", "hi", "random", "heap"])

prefix = "he"
words = [w for w in trie.all_words_beginning_with_prefix(prefix)]

print("Lazy method:\n Prefix: %s, Words: %s, Count: %d" % (prefix, words, len(words)))
print("Proactive method:\n Word count for '%s': %d" % (prefix, trie.word_count(prefix)))

输出：

Lazy method:
 Prefix: he, Words: ['hello', 'heap'], Count: 2
Proactive method:
 Word count for 'he': 2

Answer 2

我会在 trie 节点中添加一个名为 is_word 的字段，其中 is_word 仅适用于单词中的最后一个字母。就像你有单词 AND 一样，is_word 对于持有字母 D 的 trie 节点来说是真的。我会只更新 is_word 为真的节点的频率，而不是单词中的每个字母。

所以当你从一个字母迭代时，检查它是否是一个单词，如果是，停止迭代，返回计数和单词。我假设在您的迭代中您会跟踪字母，并不断将它们添加到前缀中。

你的 trie 是一个 multi-way trie。