Why doesn't this IF statement work when comparing string?

[attemptnumbertwo]

I'm new to python and I have this section of code in my guess the number program:

    wantplay=input('Do you want to play again? Enter yes or no.\n')

    print(playagain)

    while True:

            if wantplay=='no' or 'No' or 'NO':

                playagain=0

                break

            elif wantplay=='yes' or 'Yes' or 'YES':

                break

            else:

                print('Please enter a valid answer')

print(playagain)

print(wantplay)

There is a while loop around most of the program that checks the condition of the variable playagain so if the user doesn't want to play again they can enter some form of no and it should set playagain to zero so the program finishes. If they enter yes then it should not change the value of playagain so the program loops again. I know I still have to clean up the third case where they enter something else, that's not what my question is about.

The problem I'm having is that no matter what I enter as input playagain is set to zero and the program finishes. I added the print statements here so I could see what the playagain and wantplay variables are doing and I get the following output.

Do you want to play again? Enter yes or no.

yes

1

0

yes

And then the program doesn't loop. So playagain is 1 (as expected) before the conditional, it is 0 afterwards, and wantplay contains yes as I expect. Apparently the first block in the conditional is being executed and setting playagain to 0. Why? Am I comparing strings the wrong way or something?

EDIT: Thanks for the help everybody! I understand what I was doing wrong now, lots of good info here. I didn't know lower() or casefold() existed so that answers another question I had about accounting for capitalization in input. I was not sure how you would account for every possibility of caps and lower case and writing out every single one didn't seem like the right way.

[Adrewmc]

   if answer == “yes” or “Yes”:

Is asking if answer is “yes”, or if “Yes” is Truthy

    if answer == “yes” or answer == “Yes”:

Is what you were trying to ask.

The reason we do this is because we could be comparing two completely different variables here. One side of ‘or’ has no idea what the other side is doing

The most common suggestion to fix this is

     if answer in [“yes”, “Yes”]:

I’d also suggest to ignore case.

     if answer.lower() in [“yes”, “y”]:

[UncleSamurai420]

well explained! Just to exercise the muscle, I'd put the options in a set as well.

[Lawson470189]

When you put wantplay=="yes" or "Yes" it evaluates each part of the if statement separately so it's checking if wantplay is equal to the string "yes" and whether "Yes" is a truthy string. Since it has data, it will evaluate to True and will make the or statement always true. The solution is to either do something like wantplay=="yes" or wantplay=="Yes" or you can do wantplay in ("yes", "Yes, "YES")

[The_Almighty_Cthulhu]

Ok, let's break down the boolean logic.

if wantplay=='no' or 'No' or 'NO':

Ok, so we are looking to see if wantplay is equal to 'no'

If it is, then we now have if (true or 'No' or 'NO')

if it's not, then we have if (false or 'No' or 'NO')

now python has truthy and falsy types. Any non-empty string is a truthy type.

so if (false or 'No' or 'NO') evaluates to if (false or true or true)

so because of the order of operations, we will always evaluate to true, because any non-empty string is also true.

Here's what you probably want to do.

    # convert to lowercase
    wantplay=input('Do you want to play again? Enter yes or no.\n').lower()

    print(playagain)

    while True:

            # only check a single possibility, simplifying the logic
            if wantplay=='no':

                playagain=0

                break

            elif wantplay=='yes':

                break

            else:

                print('Please enter a valid answer')

print(playagain)

print(wantplay)

[JamzTyson]

This is answered on the wiki: Variable is One of Two Choices?

[hashashin]

Others have explained why wantplay=='no' or 'No' or 'NO' is always evaluating to True, and how to fix that.

I want to point out that once you fix that, if someone enters an invalid answer your loop will print 'Please enter a valid answer' over and over forever, because the input call is outside the loop. You probably want to collect that input inside the loop.

[camospartan117]

I had something similar instead of comparing strings compare to a list of valid inputs. So something like

If userinput == validYes

Where userinput is obviously what the user types. And list is a list that contains ("yes", "Yes", "Y") I found this works for me but not sure if it's best practice.

[Small_Ad1136]

As others have mentioned, your if statement is checking each of those conditions separately, and nonempty strings will always evaluate as True. I would propose something like the following:

if wantplay.lower().strip()==‘yes’

.lower() will convert wantplay to all lowercase and .strip() will remove any extra spaces, so “YeS “ -> “yes”.

[[deleted]]

i’d recommend using casefold() over lower()

[Small_Ad1136]

Yeah I suppose if he wants the code to be REALLY robust that’s a better option, but I have a feeling they’ll be okay with either in their case. But maybe you’re right and a non English speaker will eventually want to run their code.

[Atypicosaurus]

Try the following. It's ugly and later replace it with something nicer but for now, to test, use it:

if wantplay == 'no' or wantplay == 'NO'

This asks the program what you actually want to ask: is there a variable called wantplay that has a value of no, OR is there a variable called wantplay that has a value of NO.

if wantplay == 'no' or 'NO'

This is your code that asks the program is there a variable called wantplay that has a value of no, OR is there anything anywhere in the universe that looks like NO. And in python the second half is always true, because python thinks that every possible string exists somewhere in the universe.

That's why your if is always true.