Case insensitivity in "if" statement (Python)

Question

I am working on a project that at one point will ask the user a yes/no question. I currently use this code to handle such questions:

def yn():
    global finalchoice
    choice=str(raw_input("Y/N: "))
    if choice == "Y":
        finalchoice="true"
    elif choice == "y":
        finalchoice="true"
    elif choice == "N":
        finalchoice="false"
    elif choice == "n":
        finalchoice="false"
    else:
        yn()
    pass

but this seems to be quite inefficient, specifically where I have to check for both "Y" and "y" or "N" and "n" separately. I've tried:

if choice == "Y" or "y":
    finalchoice="true"

Unfortunately, all this does is ignore the 'else' command and will pass whatever I give it.

Any tips?

You can convert you input into lower case so even if user inputs upper-case if will convert it into lower case and you can use 'y' or 'n' in your if case. 'choice=str.lower(raw_input("Y/N: "))' — Tanveer Alam, Apr 24 '14 at 07:12
@Aशwiniचhaudhary this question is definitely close, but Sudipta gave the answer I was really looking for. — ExplodingKittens, Apr 24 '14 at 07:15
@TanveerAlam also offered a great solution. Thanks to the both of you! — ExplodingKittens, Apr 24 '14 at 07:16
@Sil why use `"true"` and not `True`? Also, you could `return` instead of using `global`. — jonrsharpe, Apr 24 '14 at 07:17
@jonrsharpe because I want to leave "finalchoice" as a string so that I can reuse it in the future. I don't want it to get set as a Boolean variable. — ExplodingKittens, Apr 24 '14 at 07:18
@Sil Kindly close the question by [accepting](http://stackoverflow.com/help/accepted-answer) the answer if it worked for you. — Sudipta, Apr 24 '14 at 07:21
@Sil that doesn't make sense, how is a Boolean *not* reusable? If you need a string representation later, `str(True).lower() == "true"`. — jonrsharpe, Apr 24 '14 at 07:24
@jonrsharpe I plan on using it for more than just "true" or "false" in the future is what I mean. If I try to assign a string to a Boolean, then I'll get an error. — ExplodingKittens, Apr 24 '14 at 07:34
@Sil what do you mean "assign a string to a Boolean"? Python is dynamically typed; if you later do `finalchoice = "foo"` that's fine, or you can use the existing value in the new string: `finalchoice = "value was {0}.".format(finalchoice)` — jonrsharpe, Apr 24 '14 at 07:36

Sudipta · Accepted Answer · 2014-04-24T07:17:34.727

6

if choice == "Y" or "y": is not right! It will always evaluate to True.

It essentially is (choice == "Y") or ("y"). And the latter is True because any non empty string in python logically evaluates to boolean True.

You should be doing:

if choice in ["Y","y"]:

Or,

if choice == "Y" or choice == "y":

Or you can use:

if choice.lower() == "y":
    finalchoice="true"

edited Apr 24 '14 at 07:17

answered Apr 24 '14 at 07:11

Sudipta

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score 1 · Answer 2 · answered Apr 24 '14 at 07:18

1

To add to @Sudipta's answer, one way would be to remove the case AND just take the first letter:

if choice.lower().startswith("y"):
    finalchoice="true"

answered Apr 24 '14 at 07:18

philshem

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Case insensitivity in "if" statement (Python)

2 Answers2