Is there any easier way to write this comparison operator statement?

user_5158900 0
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I'm fairly new to Python. I have the following if statement:

if(userGuess == "0" or userGuess == "1" or userGuess == "2" or userGuess == "3" or userGuess == "4" or userGuess == "5" or userGuess == "6" or userGuess == "7" or userGuess == "8" or userGuess == "9"):
    print("\n>>>error: cannot use integers\n")
    continue

Basically, the loop will reset if the user inputs any numbers. Is there any way to write this statement to make it a little more efficient? (i.e less code and cleaner)

if-statement python
3 Answers
possibilities = ["0", "1", "2", "3", "4", "5", "6", "7", "8", "9"]

if userGuess in possibilities:
    #do something

Or alternatively, if you are ok with doing a comparison with integers instead you could do the following:

if userGuess < 10:
    #do something
user_62944830

You can do this:

nums = [str(i) for i in range(10)] # gets a list of nums from 0 to 9    

if userGuess in nums:
    print("Num found")
user_124885090

Assuming userGuess is a string and can be other than a single character, 

if(any(c.isdigit() for c in userGuess):
    ....

if the guess should be exactly one character, you can 

if(len(userGuess) != 1 or userGuess in "0123456789"):
    ....

or

if(len(userGuess) != or userGuess.isdigit()):
    ...

Come to think of it, isdigit is the better way to go. Suppose the user entered the Bengali number ৩? ৩.isdigit() is True.

user_6420700
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