python 实现敏感词过滤的方法
如下所示:
#!/usr/bin/python2.6 # -*- coding: utf-8 -*- import time class Node(object): def __init__(self): self.children = None # The encode of word is UTF-8 def add_word(root,word): node = root for i in range(len(word)): if node.children == None: node.children = {} node.children[word[i]] = Node() elif word[i] not in node.children: node.children[word[i]] = Node() node = node.children[word[i]] def init(path): root = Node() fp = open(path,'r') for line in fp: line = line[0:-1] #print len(line) #print line #print type(line) add_word(root,line) fp.close() return root # The encode of word is UTF-8 # The encode of message is UTF-8 def is_contain(message, root): for i in range(len(message)): p = root j = i while (j<len(message) and p.children!=None and message[j] in p.children): p = p.children[message[j]] j = j + 1 if p.children==None: #print '---word---',message[i:j] return True return False def dfa(): print '----------------dfa-----------' root = init('/tmp/word.txt') message = '四处乱咬乱吠,吓得家中11岁的女儿躲在屋里不敢出来,直到辖区派出所民警赶到后,才将孩子从屋中救出。最后在征得主人同意后,民警和村民合力将这只发疯的狗打死' #message = '不顾' print '***message***',len(message) start_time = time.time() for i in range(1000): res = is_contain(message,root) #print res end_time = time.time() print (end_time - start_time) def is_contain2(message,word_list): for item in word_list: if message.find(item)!=-1: return True return False def normal(): print '------------normal--------------' path = '/tmp/word.txt' fp = open(path,'r') word_list = [] message = '四处乱咬乱吠,吓得家中11岁的女儿躲在屋里不敢出来,直到辖区派出所民警赶到后,才将孩子从屋中救出。最后在征得主人同意后,民警和村民合力将这只发疯的狗打死' print '***message***',len(message) for line in fp: line = line[0:-1] word_list.append(line) fp.close() print 'The count of word:',len(word_list) start_time = time.time() for i in range(1000): res = is_contain2(message,word_list) #print res end_time = time.time() print (end_time - start_time) if __name__ == '__main__': dfa() normal()
测试结果:
1) 敏感词 100个
----------------dfa----------- ***message*** 224 0.325479984283 ------------normal-------------- ***message*** 224 The count of word: 100 0.107350111008
2) 敏感词 1000 个
----------------dfa----------- ***message*** 224 0.324251890182 ------------normal-------------- ***message*** 224 The count of word: 1000 1.05939006805
从上面的实验我们可以看出,在DFA 算法只有在敏感词较多的情况下,才有意义。在百来个敏感词的情况下,甚至不如普通算法
下面从理论上推导时间复杂度,为了方便分析,首先假定消息文本是等长的,长度为lenA;每个敏感词的长度相同,长度为lenB,敏感词的个数是m。
1) DFA算法的核心是构建一棵多叉树,由于我们已经假设,敏感词的长度相同,所以树的最大深度为lenB,那么我们可以说从消息文本的某个位置(字节)开始的某个子串是否在敏感词树中,最多只用经过lenB次匹配.也就是说判断一个消息文本中是否有敏感词的时间复杂度是lenA * lenB
2) 再来看看普通做法,是使用for循环,对每一个敏感词,依次在消息文本中进行查找,假定字符串是使用KMP算法,KMP算法的时间复杂度是O(lenA + lenB)
那么对m个敏感词查找的时间复杂度是 (lenA + lenB ) * m
综上所述,DFA 算法的时间复杂度基本上是与敏感词的个数无关的。
相关推荐
XCMercy 2020-01-19
HMHYY 2020-07-28
ELEMENTS爱乐小超 2020-07-04
amazingbo 2020-06-28
alicelmx 2020-06-16
minkee 2020-06-09
逍遥友 2020-06-02
嗡汤圆 2020-05-10
whbing 2020-05-05
zhuxianfeng 2020-05-02
assastor 2020-05-01
JessePinkmen 2020-05-01
hongxiangping 2020-04-30
theta = np.zeros #theta = array,构造全为零的行向量。grad[0,j] = np.sum/len #∑term / m. return value > threshol
Kwong 2020-04-26
88483063 2020-04-23
xirongxudlut 2020-04-19