Rock Paper Scissor GAME

[ichabod801]

As a third way to do it, I like this dictionary for rock, paper, scissors:

wins = {'rock': 'scissors', 'scissors': 'paper', 'paper': 'rock'}

It stores what beat what in a slightly different data structure than perfringo used. Again, checking for the win is very simple:

if user_choice == computer_choice:
    # draw
elif wins[user_choice] == computer_choice:
    # user wins
else:
    # computer wins

Note that I like this because it scales well for expanding to rock-paper-scissors-lizard-spock:

wins = {'rock': ['scissors', 'lizard'], 'scissors': ['paper', 'lizard'], 'paper': ['rock', 'spock'],
     'lizard': ['paper', 'spock'], 'spock': ['scissors', 'rock']}

You just need to change to testing with the in operator:

if user_choice == computer_choice:
    # draw
elif computer_choice in wins[user_choice]:
    # user wins
else:
    # computer wins

The lesson being that there may be two different algorithms/data structures that are basically equivalent for the problem at hand, but the choice of which to use may be driven by other parts of the application.