Procedural style Rock Paper Scissors game using Python 3

Question

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()

Answer 1

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.

• Le Quynh Anh: 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?
• Sometowngeek: I'm just using it as an example. It's another way of saying "If a variable holds one of these values."

Answer 2

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!

• Le Quynh Anh: 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.
• Sometowngeek: @BlahBlah, In the beats() method, there's a possibility of tie. Might want to add something to handle that. :)
• Isi: @Sometowngeek, I believe I handled it in the play() function. if decision1 == decision2 is handling the tie. :)

Answer 3

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.