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

Question

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)

Try `if userGuess.isnumeric():` - see [`str.isnumeric()`](https://docs.python.org/3.8/library/stdtypes.html#str.isnumeric). — metatoaster, Jun 09 '20 at 03:57
As a note on examples, `continue` can't be used outside of a loop. Its best to post a running example. In your case, assigning a value to `userGuess` and leaving out the `continue`. — tdelaney, Jun 09 '20 at 04:07
thanks for all the help. i didn't know you can int() with a str that's a number — awoldt, Jun 09 '20 at 04:11
If `userGuess` is any string of digits greater than 0 length, then `int` works. Suppose its `User1234`, `int` would fail, but the guess would contain digits. I don't know what you want in that case. — tdelaney, Jun 09 '20 at 04:15

score 3 · Accepted Answer · answered Jun 09 '20 at 03:56

3

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

answered Jun 09 '20 at 03:56

chumbaloo

671
6
16

could you explain how the loop if userGuess in possibilities: works? – awoldt Jun 09 '20 at 04:25
Sure, no problems. The if statement is followed by a conditional check (which if evaluated to true, executes the next indented block of code). The use of the keyword in, is best explained already by this answer (https://stackoverflow.com/a/38204487). It simply checks if the value of userGuess is inside the list of items in possibilities. – chumbaloo Jun 09 '20 at 04:29

score 3 · Answer 2 · edited Jun 09 '20 at 04:02

3

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")

edited Jun 09 '20 at 04:02

chumbaloo

671
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16

answered Jun 09 '20 at 03:59

bhristov

3,137
2
10
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tdelaney · Answer 3 · 2020-06-09T04:12:48.423

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.

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

3 Answers3