检查字符串是否包含列表中任何单词的最快方法 [英] Fastest way to check does string contain any word from list

问题描述

我有Python应用程序.

I have Python application.

有450个禁止的短语列表.有来自用户的消息.我想检查一下，此消息是否包含任何这种禁止的言词.最快的方法是什么?

There is list of 450 prohibited phrases. There is message got from user. I want to check, does this message contain any of this prohibited pharases. What is the fastest way to do that?

当前我有以下代码:

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).

Is there any faster way to do that? I need to handle message (max 500 chars) in less than 1 second.

推荐答案

有 any 专用于此的内置函数:

There is the any built-in function specially for that:

>>> 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:

---

Alternatively you could check the intersection of the sets:

>>> 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:

But you won't be able to check for subwords:

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

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