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I am a newbie to Python and am trying to do the Rock, Paper and Scissors game to revise my knowledge that I have earned so far.

Overview of the program:

  • This is a two-player game
  • Ask for player plays (including name, their choice: rock or paper or scissors using input)
  • Print out a message of winner
  • Ask if the user want to start a new game

This code is written in procedural style.

Can anyone show me some weaknesses in my code and how I can improve on them?

def beats(a, b): 
    tup = ('rock', 'paper', 'scissors')
    if (a == tup[1] and b == tup[2]) or (a == tup[2] and b == tup[1]):
        return tup[2]
    elif (a == tup[1] and b == tup[0]) or (a == tup[0] and b == tup[1]):
        return tup[1]
    elif (a == tup[0] and b == tup[2]) or (a == tup[2] and b == tup[0]):
        return tup[0]


def players_input():
    name = input('What is your name?: ')
    while True:
        decision = input('Paper, Rock, or Scissors?: ')
        if decision == 'rock' or decision == 'paper' or decision == 'scissors':
            break
        else:
            print("Invalid choice! Try again!")
            continue
    return '{} chooses {}'.format(name, decision)


def play():
    player1 = players_input()
    player2 = players_input()
    print('{}\n{}'.format(player1, player2))
    name1, verb, decision1 = player1.split(" ")
    name2, verb, decision2 = player2.split(" ")
    returnvalue = beats(decision1.lower(), decision2.lower())
    if decision1.lower() == returnvalue and decision2.lower() != returnvalue:
        print('{} wins'.format(name1))
    elif decision2.lower() == returnvalue and decision1.lower() != returnvalue:
        print('{} wins'.format(name2))
    else:
        print('It\'s a tie')


def play_again():
    while True:
        choice = input('Do you want to continue?(Y/N): ')
        if choice.upper() == "Y":
            play()
        elif choice.upper() == "N":
            print("Game over")
            break
        else:
            print("Invalid Input! Try again!")


play()
play_again()
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3 Answers 3

4
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I recommend simplifying:

if decision == 'rock' or decision == 'paper' or decision == 'scissors':

to:

if decision.lower() in ('rock', 'paper', 'scissors'):

This will check whether the decision exists in a given tuple.

This helps simplifying the idea "If a variable contains one of these values".

Another example is "If variable x contains either 1, 3, or 5." Instead of coding it like this:

if x == 5 or x == 3 or x == 1

It can be simplified into:

if x in (1, 3, 5)

See this Gist to see my suggestion as a whole.

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2
  • \$\begingroup\$ I do not catch the idea of if x == 5 or x == 3 or x == 1? What is x, 1, 3 and 5? Can you help me explain it? \$\endgroup\$ Commented Jul 4, 2017 at 18:14
  • \$\begingroup\$ I'm just using it as an example. It's another way of saying "If a variable holds one of these values." \$\endgroup\$ Commented Jul 4, 2017 at 18:21
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First of all, it is good you are reviewing your knowledge.
Secondly, I would like to add to Sometowngeek's suggestion. Implement his one.

Thirdly, instead of writing:

name1, verb, decision1 = player1.split(" ")

you could write:

name1, decision1 = player1.split(" chooses ")

I made some changes to your play() function:

def play():
    player1 = players_input()
    player2 = players_input()

    print('{}\n{}'.format(player1, player2))

    name1, decision1 = player1.split(" chooses ")
    name2, decision2 = player2.split(" chooses ")

    if decision1 == decision2:
        print('It\'s a tie')
    else:
        returnValue = beats(decision1, name1, decision2, name2)
        print('{} wins'.format(returnValue))

And some more changes to your beats() function:

def beats(a, name1, b, name2):
    tup = ('rock', 'paper', 'scissors')
    if (a == tup[0] and b == tup[2]) or (a == tup[1] and b == tup[0]) or (a == tup[2] and b == tup[1]):
        return name1
    else:
        return name2

In beats() function, it is better to return the names of the players; so, you will not use if in your play() function.

Finally, good luck with your future coding!

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  • \$\begingroup\$ I had thought that the split function can only be passed in a delimeter (like space, -, so on). Thank you, I have realized lots of things. \$\endgroup\$ Commented Jul 6, 2017 at 8:19
  • \$\begingroup\$ @BlahBlah, In the beats() method, there's a possibility of tie. Might want to add something to handle that. :) \$\endgroup\$ Commented Jul 6, 2017 at 20:45
  • 2
    \$\begingroup\$ @Sometowngeek, I believe I handled it in the play() function. if decision1 == decision2 is handling the tie. :) \$\endgroup\$
    – Isi
    Commented Jul 7, 2017 at 9:38
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name = input("name?")
def players_input(player):
    while True:
    decision = input('Paper, Rock, or Scissors?: ')
    if decision == 'rock' or decision == 'paper' or decision == 'scissors':
        break
    else:
        print("Invalid choice! Try again!")
        continue
return '{} chooses {}'.format(player, decision)

So you won't be prompted to input your name every time you play again.

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