Python - If string contains a word from a list or set

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.

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!")

Result:

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!

Using Regular Expression:

Pattern used is 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.

You can use sets, if only you want to check the existance of swear words,

a_swear_set = set(Swear)

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