Rock, Paper Scissors game in Python

Question

I wrote a simple Rock, Paper Scissors game in Python for class. As a beginner, I would like some input as to how I could abstract or make my code smoother. I feel like the repeated use of if or elif statements are overbearing. I would like any tips or advice, or perhaps a different method of approach.

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)
 choice = comp
 # assign (int) to user choices
 if user == 'r':
 user = 1
 elif user == 'p':
 user = 2
 elif user == 's':
 user = 3
 # assign string to computer random choice
 if comp == 1:
 choice = 'rock'
 elif comp == 2:
 choice = 'paper'
 elif comp == 3:
 choice = 'scissors'
 print('Computer picked:', choice)
 # compare selection to figure out the winner
 if user == 1 and comp == 3:
 print('You win against the computer\n')
 elif user == 3 and comp == 1:
 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')

• the scion: What about 'switch case'?
• Martijn Pieters: @thescion: As this is Python, there is no switch case. That said, RPS is trivially coded up with a dictionary.
• Mast: We have a rock-paper-sciccors tag. Feel free to browse those questions for alternative implementations. There should be a couple in python as well.
• Toby Speight: Your title should indicate what the code is intended to do - please read our How to Ask, because it's quite different to that of Stack Overflow.
• TylerH: I don't think scissors made of paper would be very effective.

Answer 1

The way you are trying to solve this problem does not seem pythonic. Since switch statements don't exist in python, you can simulate switch-like behaviour using dictionaries.

For example, instead of using if-else to set the value of user, you can so something like

user_options = {
 'r': 1,
 'p': 2,
 's': 3
}
user = user_options[user]

Since functions and classes are first-class objects in python, if you want to accomplish some less trivial, you can use them too:

def func1():
 # do some operation
def func2():
 # some other operation
def default_func():
 # some default action
options = {
 '1': func1,
 '2': func2,
 'default': default_func
}
value = raw_input('enter a number')
result = options.get(int(value), options.get('default'))()

If your functions are simple enough, you can use lambdas as values, which would definitely be more concise.

• Richard Peña: What is user = user_options[user] doing? You define the dictionary first and then set user equal to the corresponding value in the dictionary?
• hspandher: In your code, you are setting user =1 if it is equal to 'r', that's exactly what that statement is doing
• wizzwizz4: Your code won't work. options[int('1')] won't lookup on '1'.
• hspandher: You are right. One has to convert into an integer first. Anyway I was just giving him only a general idea on how to approach toward solution.

Answer 2

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')

• Martin Ueding: Why do you have curly braces instead of square brackets around the list of win-conditions? That makes it a dict without values but only keys?
• mkrieger1: @MartinUeding That's a set.
• Martin Ueding: Cool, I did not know that. Seems to be added in Python 3.1, so it is safe to use on all my targeted platforms. Lookup in the set will be faster, so that is a neater solution; though it won't matter with three elements.
• Zefick: That makes a Set. Technically it is indeed a dictionary with only keys. This structure from Python 2, since Python 2.7, {} can be used to declare a set

Answer 3

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. 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 = (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 an 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}
}

• Dubu: You could probably replace your "magic" 3 with len(options).
• NonlinearFruit: @Dubu The magic number works for 1 option and 3 options but fails for 5 and larger. Changing to len(options) won't improve this (and it will make it less readable). But abstracting away the 3 is a good idea. I added an 'advanced check' section which contains a generic solution for any odd number of options.
• Richard Peña: Is 'r':{'name':'Rock','value':1} a dictionary, within a dictionary? options = { } is the "parent" dictionary?
• NonlinearFruit: @RichardPeña Yes, it is a dictionary in a dictionary with options being the parent. It makes the code a little more readable.
• NonlinearFruit: @RichardPeña Thanks for catching that. In Python 3, we need to make the keys into a list: random.choice(list(options.keys())).

Answer 4

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')
   

• MKesper: 'if choices.find(user) == -1': I find it more readable to write 'if user not in choices'
• Fallen: agreed and updated :)
• MKesper: Your indentation seems to have become corrupted.

Answer 5

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.

• ferada: Welcome to Code Review, your first answer looks good, enjoy your stay!
• Richard Peña: How does if (user_choice, comp_choice) in {(1, 3), (2, 1), (3, 2)}: work? If user is: 1 while comp is 3, 2 while comp is 1, 3 while comp is 2?
• wizzwizz4: @RichardPeña In that case, the other option (else) will be chosen, because the other person wins.
• jthulhu: @RichardPeña you create an object of type tuple (user_choice, comp_choice), and check if there is the same in a set that contains 3 tuples. The difference between set and tuples is that each object in set is unique, and set is not ordered. This increases efficiencity.

Answer 6

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')