I am trying to get user input like easy or e equal to E, to choose game modes.

I have tried doing:

difficulty_chooser = input("Welcome to the number game! Choose your difficulty!\n(E)asy, (M)edium, (H)ard: ").lower()

to make it so that if the user types e instead of E, it will show the same output as if E was typed. This does not do anything whatsoever, as when I type e, or any other difficulty modes in lowercase, the code ends.

The code is down below.

def intro():

    difficulty_chooser = input(
        "Welcome to the number game! Choose your difficulty!\n(E)asy, (M)edium, (H)ard: "
    )

    if difficulty_chooser == "E":
        print("Time for easy mode!")
        easy_game()

    if difficulty_chooser == "M":
        print("Time for a bit of challenge in Medium mode!")
        medium_game()

    if difficulty_chooser == "H":
        print("Time for the biggest challenge!")
        hard_game()

CodePudding user response：

Using .upper() will solve the issue, but I figure it's also worth pointing out that if you're trying to handle edge cases then you ought to handle:

1. The case when someone enters something other than E, M, or H (case insensitive).
2. The fact that all three difficulties will execute sequentially, since you're using if as opposed to elif statements.

Fixing these issues, we get the following code:

def intro():
    valid_inputs = {'E', 'M', 'H'}
    difficulty_chooser = None
    while difficulty_chooser not in valid_inputs:
        difficulty_chooser = input(
            "Welcome to the number game! Choose your difficulty!\n(E)asy, (M)edium, (H)ard: "
        ).upper()
        
        if difficulty_chooser not in valid_inputs:
            print("Please enter 'E', 'M', or 'H'.")

    if difficulty_chooser == "E":
        print("Time for easy mode!")
        easy_game()
    elif difficulty_chooser == "M":
        print("Time for a bit of challenge in Medium mode!")
        medium_game()
    else:
        print("Time for the biggest challenge!")
        hard_game()

which should be more robust against malformed inputs.

CodePudding user response：

Try to utilize a mapper instead of if-else.

easy_game = medium_game = hard_game = lambda: print('facked starter')

def intro():
    choosen_mapper = {'E': easy_game, 'M': medium_game, 'H': hard_game}
    prompt_mapper = {'E': "Time for easy mode!",
                     'M': "Time for a bit of challenge in Medium mode!", 
                     'H':"Time for the biggest challenge!"}
    assert choosen_mapper.keys() == prompt_mapper.keys()

    difficulty_chooser = None
    while difficulty_chooser not in choosen_mapper:
        difficulty_chooser = input(
            "Welcome to the number game! Choose your difficulty!\n(E)asy, (M)edium, (H)ard: "
        ).upper()
    print(prompt_mapper[difficulty_chooser])
    choosen_mapper[difficulty_chooser]()