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字典树用于单词联想
最近要做一个单词联想的功能,经过调研选择使用字典树,节省空间,查找快。
贴上代码
class Trie(object): def __init__(self): self.path = {} self.value = None self.valid = False def __setitem__(self, key, val): head = key[0] if head in self.path: node = self.path[head] else: node = Trie() self.path[head] = node if len(key) > 1: remains = key[1:] node.__setitem__(remains, val) else: node.value = val node.valid = True def __getitem__(self, key): head = key[0] if head not in self.path: raise KeyError(key) else: node = self.path[head] if len(key) > 1: remains = key[1:] try: return node.__getitem__(remains) except KeyError: raise KeyError(key) elif node.valid: return node.value else: raise KeyError(key) def get(self, key, default=None): try: return self.__getitem__(key) except KeyError: return default def __contains__(self, key): try: self.__getitem__(key) return True except KeyError: return False def __keys__(self, prefix=[], seen=[]): result = [] if self.valid: is_str = True val = ‘‘ for k in seen: if type(k) != str or len(k) > 1: is_str = False break val += k if is_str: result.append(val) else: result.append(prefix) if len(prefix) > 0: head = prefix[0] prefix = prefix[1:] if head in self.path: nextpaths = [head] else: nextpaths = [] else: nextpaths = self.path.keys() for key in nextpaths: nextseen = [] nextseen.extend(seen) nextseen.append(key) result += self.path[key].__keys__(prefix, nextseen) return result def keys(self, prefix=[]): return self.__keys__(prefix) def __iter__(self): for item in self.keys(): yield item raise StopIteration trie = Trie() trie[‘abc‘] = 1 trie[‘salt‘] = 1 trie[‘hello‘] = 1 trie[‘alter‘] = 1 print trie[‘abc‘] print trie.keys() print trie.keys(‘a‘) for key in trie: print key
>> 1
>> [‘abc‘, ‘alter‘, ‘hello‘, ‘salt‘]
>> [‘abc‘, ‘alter‘]
>> abc
>> alter
>> hello
>> salt
代码参考 https://github.com/bdimmick/python-trie/blob/master/trie.py
字典树用于单词联想
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