最快的检查方式是字符串包含列表中的任何单词

Question

我有Python应用程序.

有450个禁止短语列表.有来自用户的消息.我想检查,这条消息是否包含任何这种禁止的删除症状.最快的方法是什么？

目前我有这个代码:

message = "sometext"
lista = ["a","b","c"]

isContaining = false

for a, member in enumerate(lista):
 if message.contains(lista[a]):
  isContaining = true
  break

有没有更快的方法呢？我需要在不到1秒的时间内处理消息(最多500个字符).

Answer 1

有专门针对它的任何内置功能:

>>> message = "sometext"
>>> lista = ["a","b","c"]
>>> any(a in message for a in lista)
False
>>> lista = ["a","b","e"]
>>> any(a in message for a in lista)
True

或者,您可以检查集合的交集:

>>> lista = ["a","b","c"]
>>> set(message) & set(lista)
set([])
>>> lista = ["a","b","e"]
>>> set(message) & set(lista)
set(['e'])
>>> set(['test','sentence'])&set(['this','is','my','sentence'])
set(['sentence'])

但是您将无法检查子词:

>>> set(['test','sentence'])&set(['this is my sentence'])