Python-如果字符串包含列表中的单词或

Question

我已经找得很彻底，没有找到合适的答案。我对Python/Programming还不熟悉，所以我非常感谢您给我的建议：

我试图搜索用户输入字符串中的某些关键字。例如，我们会说过滤掉亵渎。从我的研究来看，我已经能够做出以下虚拟的例子：

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!")

使用上述方法，如果用户将整个字符串中的一个确切的单词输入整串，它将打印“退出诅咒”；但是，如果用户输入“诅咒”或“我喜欢说诅咒”，它会打印“听起来太好了！”

最终，我需要的是能够搜索整个字符串中的关键字，而不是整个字符串的完全匹配。例如：“我去了公园，觉得尖叫的诅咒应该会在比赛中变成现实。

Answer 1

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 2

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 3

你可以使用集合，如果你想检查脏话的存在

a_swear_set = set(Swear)

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