Rock, Paper Scissors game in Python

In Python you can compare multiple values the following way:

if (user, comp) in {(0, 2), (1, 0), (2, 1)} :
    print('You win against the computer\n')
elif user == comp :
    print('You tie against the computer\n')
else :
    print('You lose against the computer\n')

Here are 3 corrections. Instead of manually converting r,s,p and 1,2,3, use a dictionary. Instead of using a random number for the computers choice, use the players options. Instead of having extra win/lose/tie checks, have one per outcome.

Fix Conversion

This is a simple way to convert r,p,s into a name or a value:

• options['r']['name'] == 'Rock'
• options['r']['value'] == 1

options = {
    'r':{'name':'Rock','value':1},
    'p':{'name':'Paper','value':2},
    's':{'name':'Scissors','value':3}
}

Fix Computer's Choice

By choosing a random key from options prevents having to hard-code values. It also allows us to lookup the name and value for the computer's choice. compChoice will be r, p, or s since those are options' keys.

compChoice = random.choice(list(options.keys()))

Fix win/lose/tie Checks

A tie only happens if compValue==userValue. Now consider +1 mod 3, for any value if you "+1 mod 3" you will get the only value that can beat it:

• rocVal+1 %3 == papVal
• papVal+1 %3 == sciVal
• sciVal+1 %3 == rocVal

compVal = options[compChoice]['value']
userVal = options[userChoice]['value']
result = 'win'

if userVal == compVal: result = 'tie'
elif userVal+1 % 3 == compVal: result = 'lose'
# else: result = 'win'

print('You '+result+' against the computer!\n')

Advanced Check

But what if you want to play Rock-Paper-Scissors-Lizard-Spock? Or a any version with more than 3 options? If that is the case, here is a scalable solution from sch. Note that the number of options should always be odd. This way each element has the same number of superiors and inferiors.

Explanation: the key piece here is the decider. We want the difference between the two choices to be greater than or equal to 0 so we have to add numOpns to it. Now picture these options as being on a circle (e.g. an analog clock). decider is the clockwise distance from b to a. [Note: if this distance is even, then the distance from a to b is odd] For any particular option, half of the remaining options are an even distance and half are an odd distance. Here we arbitrary choose that odd distances correspond to a loss.

compVal = options[compChoice]['value']
userVal = options[userChoice]['value']
numOpns = len(options)
decider = (numOpns+userVal-compVal) % numOpns
result = 'win'

if decider == 0: result = 'tie'
elif decider%2 == 0: result = 'lose'
# else decider%2 == 1: result = 'win'

print('You '+result+' against the computer!\n')

In this case, hints are definitely a problem if hard coded. So here is a fix for that:

for k,v in options.iteritems():
    print(k.upper()+" or "+k+" for "+v['name'])
print('Q or q to quit')

And while we are at it, here is a options dictionary that simulates the correct Rock-Paper-Scissors-Lizard-Spock relationship.

options = {
    'r':{'name':'Rock','value':5},
    'l':{'name':'Lizard','value':4},
    'o':{'name':'Spock','value':3},
    's':{'name':'Scissors','value':2},
    'p':{'name':'Paper','value':1}
}

My second pass,

Improvements:

1. Use list and string to find the user's and computer's move easily, instead of doing if-elses.

2. Never trust user input. Verify that the user input is one of r, p, s or q.

   import random
   
   user = ''
   while user != 'q':
       print('Enter your choice')
       print('R or r for rock\nP or p for paper\nS or s for scissors\nQ or q to quit')
       user = input('')
       user = user.lower()
       choices = 'rps'
   
       if user == 'q':
           break
       if user not in choices:
           print 'invalid choice'
           continue
       user = choices.find(user)
   
       comp = random.randrange(0, 3)
       choice = comp
   
       choices = ['rock', 'paper', 'scissors']
       print('Computer picked:', move_choices[comp])
   
       # compare selection to figure out the winner
       if user == 0 and comp == 2:
           print('You win against the computer\n')
       elif user == 2 and comp == 0:
           print('You lose against the computer\n')
       elif user > comp:
           print('You win against the computer\n')
       elif user < comp:
           print('You lose against the computer\n')
       elif user == comp:
           print('You tie against the computer\n')
   

Apart from the if/elif improvements, you can also to some other things (here I take tips from all the answers to give you a good solution):

import random

print('Tip:\nR or r for rock\nP or p for paper\nS or s for scissors\nQ or q to quit')
while True:
    user = input('Enter you choice\n> ').lower()
    options = {'r': 1, 'p': 2, 's': 3}
    comp_trans = {1: 'rock', 2: 'paper', 3: 'scissors'}
    if user == 'q':
        break
    elif user not in options.keys():
        print('Invalid input, please re try')
        continue
    comp_choice = random.randint(1, 3)
    user_choice = options[user]
    print('Computer choosed:', comp_trans[comp_choice])
    if (user_choice, comp_choice) in {(1, 3), (2, 1), (3, 2)}:
        print('You won!')
    elif user_choice == comp_choice:
        print('You tied!')
    else:
        print('You lost!')

Changes:

1) Instead of giving tips every game, we give it once.

2) We dont need to check if user == 'q' each loop (in while statement) because you already do it in the loop

3) if you need only once a value, instead of creating a variable for it and using the variable directly use the value (in displaying the computer choose)

4) we need to check the input, or we may get a KeyError on the line

user_choice = options[user]

5) you asked what does the previous line do: it defines the variable user_choice to the value options[user]. options[user] returns the value in the dictionary options that has the key user. For example, if user = 'r', options[user] will return 1.

Python has native lists, tuples and dictionaries (and a number of other data-structures). If you use the builtin data structures wisely you can often drastically simplify your code. In your case you are mapping back and forth between ints (1,2,3), characters ('r', 'p' and 's') and names ('rock',.. etc). Having the ints isn't necessary and the mapping from chars to names is better done with a dict.

import random

options = {"r": "rock", "p": "paper", "s": "scissors"}

while True:
    user = input('Choose r for rock, p for paper, s for scissors or q to quit: ')
    user = user.lower()

    if user == 'q':
        break

    if user not in options.keys():
        continue

    choice = random.choice(list(options.keys()))
    print('Computer picked:', options[choice])

    if choice == user:
        print('You tie against the computer\n')
    elif (user, choice) in (("r", "s"), ("p", "r"), ("s", "p")):
        print('You win against the computer\n')
    else:
        print('You lose against the computer\n')

Here, Dictionaries are used to make the selection of user and comp. This may remove the unnecessary comparisons.And the conditions which may lead to same results (win or loss) is being combined so it may compact the code.

import random
user = ''

while user != 'q':
    print('Enter your choice')
    print('R or r for rock\nP or p for paper\nS or s for scissors\nQ or q to quit')
    user = input('')
    user = user.lower()
    if user == 'q':
        break
    comp = random.randrange(1, 4)
    user_dict={'r':1,'p':2,'s':3}
    user=user_dict[user]
    comp_dict={1:'rock',2:'paper',3:'scissors'}
    choice=comp_dict[comp]

    print('Computer picked:', choice)

    # compare selection to figure out the winner
    if (user == 1 and comp == 3) or (user > comp):
        print('You win against the computer\n')
    elif (user == 3 and comp == 1) or (user < comp):
        print('You lose against the computer\n')
    elif user == comp:
        print('You tie against the computer\n')