Question

我搜索得非常彻底，但没有找到合适的答案。我是Python / Programming的新手，所以我很感激我能得到的任何建议：

我正在尝试搜索某些关键字的用户输入字符串。例如，我们会说过滤亵渎。根据我的研究，我已经能够制作以下虚拟示例：

Swear = ("curse", "curse", "curse") #Obviously not typing actual swear words, created a set
Userinput = str.lower(input("Tell me about your day: "))

if Userinput in Swear:
     print("Quit Cursing!")
else:
     print("That sounds great!")

使用上述内容，如果用户输入一个确切的单词形式作为整个字符串，它将打印“quit cursing”;但是，如果用户输入“诅咒”或“我喜欢说诅咒”，它将打印“听起来很棒！”

最终我需要的是能够在整个字符串中搜索关键字，而不是整个字符串的完全匹配。例如：“我去了公园，感觉像尖叫诅咒 s”应该返回真实的比赛。

Answer 1

Swear = ["curse", "curse", "curse"]

for i in Swear:
    if i in Userinput:
        print 'Quit Cursing!'

You should read up on the differences between lists and tuples.

Answer 2

Swear = ("curse", "curse", "curse") 
Userinput = str.lower(raw_input("Tell me about your day: "))

if any(Userinput.find(s)>=0 for s in Swear):
     print("Quit Cursing!")
else:
     print("That sounds great!")

结果：

Tell me about your day: curse
Quit Cursing!

Tell me about your day: cursing
That sounds great!

Tell me about your day: curses
Quit Cursing!

Tell me about your day: I like curse
Quit Cursing!

使用正则表达式：

使用的模式是r"\bcurse[\w]*"。

Swear = ("curse", "curse", "curse") 
Userinput = str.lower(raw_input("Tell me about your day: "))

if any(match.group() for match in re.finditer(r"\bcurse[\w]*", Userinput)) :
     print("Quit Cursing!")
else:
     print("That sounds great!")

finditer(pattern, string, flags=0)
    Return an iterator over all non-overlapping matches in the
    string.  For each match, the iterator returns a match object.

    Empty matches are included in the result.

Answer 3

你可以使用集合，只要你想检查咒骂词的存在，

a_swear_set = set(Swear)

if a_swear_set & set(Userinput.split()):
     print("Quit Cursing!")
else:
     print("That sounds great!")

Python - 如果string包含列表或集合中的单词

3 个答案:

结果：