Play again, in a guessing game

[banidjamali]

Hello!

Here the computer comes up with a number in range(1, 9), and we try to guess it, until we come up with the correct number.
We can exit anytime during the game and also at the end we have the choice to play again.

import random
def guessing():
    computer = random.randint(1, 9)
    number_of_guesses = 0
    while True:
        number_of_guesses += 1
        pick = input("Guess, if you dare: ")
        if pick.lower() == "exit":
            print("Better luck next time!")
            print("Attempts:", number_of_guesses - 1)
            break
        elif int(pick) > computer:
            print("Too high.")
        elif int(pick) < computer:
            print("Too low.")
        elif int(pick) == computer:
            print("Yes!")
            if number_of_guesses == 1:
                print("Are you God?!")
            elif number_of_guesses < 3:
                print(f"Impressive! Only {number_of_guesses} attemps.")
            else:
                print(f"Finally! After {number_of_guesses} attemps.")
            break
    again()

def again():
    if input("Continue?(y/n) ") == 'y':
        guessing()
    else:
        exit

guessing()

I've been told that I don't need to define a separate function for just asking the user if they want to play again. But I can't seem combine the two functions together.

Thank you for reading.

[buran]

Yes, it doesn't make sense to make a separate function.
the problem is that you need to call sys.exit(). Just exit doesn't do anything even if you have imported it from sys (not shown here).
Anyway, calling recursively the game function is not good idea. Ultimately (although unlikely) you will reach max recursive calls limit.

If it has to be in function - better to return user choice (True/False)

[deanhystad]

All in one.

while True:
    # Start a game
    while True:
        # Play game
        number_of_guesses += 1
        ...
        elif int(pick) == computer:
            ...
            break
    # Do you want to play again?
    if input('Continue (y/n)? ') != 'y':
        break;

You can do this with just one function though I think it cleaner to write a function that plays the game and repeatedly call the game function if you want to play again. I would implement the logic like this:

def play_game():
    """play the game"""

while True:
    play_game()
    if input('Play again (y/n)? ') not in ('Y','y'):
        break

The way you are doing it, with recursion, is ugly. It mixes up the game function with the repeat logic. The game should play the game and the repeat code should ask if you want to play a again and call the game function or end the loop. Recursion also causes a problem where you have to use the nuclear option of exiting the program to unwind the recursion stack.

Both choices have only one function.

[snippsat]

A function more may not be needed as buran say or deanhystad show.
For me it make better/cleaner to have repeat game and quit game in one function and the game logic in one function.
Then always fall back to menu function where can start a new game or quit out.

import random

def play_game():
    print("Guess a number betwwen 1 and 10")
    secret_number = random.randint(1, 10)
    guess, tries = 0, 0
    while guess != secret_number:
        guess = int(input("Take a guess: "))
        if guess > secret_number:
            print ("Lower")
        elif guess < secret_number:
            print ("Higher")
        tries += 1
    print(f'You guessed it! The number was {guess} in {tries} tries\n')

def menu():
   while True:
       print('(1) Play guess number game')
       print('(Q) Quit')
       choice = input('Enter your choice: ').lower()
       if choice == '1':
           play_game()
       elif choice == 'q':
           return False
       else:
           print(f'Not a correct choice: {choice}')

if __name__ == '__main__':
    menu()

[banidjamali]

def menu():

Very helpful. Thank you.