Question

我必须编写类似于石头剪刀的程序游戏，但有五个选项而不是三个。我能够使用ifs系统编写代码，但我想知道是否有更好的方法来编写代码。

游戏规则：

如您所见，共有5个选项（X→Y表示X胜过Y）：

Rock→Lizard＆amp;剪刀
纸张→Rock＆amp; Spock
剪刀→纸张＆amp;蜥蜴
Lizard→Spock＆amp;论文
Spock→剪刀＆amp;摇滚

主要代码：

import random
from ex2_rpsls_helper import get_selection

def rpsls_game():
  com_score = 0
  player_score = 0
  draws = 0
  while(abs(com_score - player_score) < 2):
    print("    Please enter your selection: 1 (Rock), 2 (Paper), 3 (Scissors), 4 (Lizard) or 5 (Spock): ")
    selection = int(input())
    # a while loop to make sure input i between 0<x<6
    while(selection <= 0 or selection > 5):
        print(    "Please select one of the available options.\n")
        selection = int(input())
    com_selection = random.randint(1,5)
    print("    Player has selected: "+get_selection(selection)+".")
    print("    Computer has selected: "+get_selection(com_selection)+".")

    # A set of else and elseif to determin who is the winner
    if(give_winner(selection, com_selection)):
        print("    The winner for this round is: Player\n")
        player_score += 1
    elif(give_winner(com_selection,selection)):
        print("    The winner for this round is: Computer\n")
        com_score += 1
    else:
        print("    This round was drawn\n")
        draws += 1

    print("Game score: Player "+str(player_score)+", Computer "+str(com_score)+", draws "+str(draws))
  if(player_score > com_score):
    return 1
  else:
    return -1

IFS系统：

def give_winner(first_selection, second_selection):
    if(first_selection is 1):
        if(second_selection is 3 or second_selection is 4):
            return True
    elif(first_selection is 2):
        if(second_selection is 1 or second_selection is 5):
            return True
    elif(first_selection is 3):
        if(second_selection is 2 or second_selection is 4):
            return True
    elif(first_selection is 4):
        if(second_selection is 2 or  second_selection is 5):
            return True
    elif(first_selection is 5):
        if(second_selection is 3 or second_selection is 1):
            return True
    return False

有什么想法吗？

Answer 1

您可以拥有if元组的列表或字典，而不是一系列复杂的(first, second)语句，

a = [(1,3), (1,4), (2,1), (2,5) ...]

def give_winner(first_selection, second_selection):
    return (first_selection, second_selection) in a

您还可以使用frozenset来提高效果。

Answer 2

你可以使用字典。

dictionary = {
    1: [3, 4],
    2: [1, 5],
    3: [2, 4],
    4: [2, 5],
    5: [3, 1]
}

def give_winner(first_selection, second_selection):
    if dictionary.has_key(first_selection):
        if second_selection in dictionary[first_selection]:
            return True
    return False

Answer 3

您可以在python中使用类。例如，您可以将播放器设为具有以下属性的类： -

    Score
    Name
    OptionChosen

等同样，你可以制作像

这样的方法

    UpdateScore()
    DeclareWinner()

等。这样你的程序就会感觉更“整洁”。您还可以创建一个包含

的main（）函数

    while True:

并将所有内容放在那里。对于前

    class Player:
        def __init__(self,name, score = 0):
          self.name = name
          self.score = score  # initially score is zero
        def ChooseOption(self, name):
           if name == "computer":
            # select choice randomly code
           else:
             var = int(input("Enter choice: "))
        def UpdateScore(self):
            self.score += 1
    def main():
        player1 = Player("Name")
        player2 = Player("Computer") 
        while True:
         resp1 = player1.ChooseOption()
         resp2 = player2.ChooseOption()
         # add other code to manipulate resp1 and resp2 here

同样你可以编写其他东西，希望这会给你一些想法

Answer 4

也为获胜者提供替代方案：

def give_winner(first_selection, second_selection):
    rules = {
        1: lambda x: x in (3, 4),
        2: lambda x: x in (1, 5),
        3: lambda x: x in (2, 4),
        4: lambda x: x in (2, 5),
        5: lambda x: x in (3, 1)
    }
    return rules[first_selection](second_selection)

Answer 5

我很喜欢建立我自己的小版Rock Paper Scissor Lizard Spock。

您通过解释som规则开始了您的帖子。所以我想，为什么不在规则中加入规则。我想用真实的单词代替数字，因为它更容易理解。但我同意每次拼写剪刀都很麻烦，所以我希望数字也可以作为输入。

from random import randint

# ex. scissors and lizard is beaten by rock
beaten_by = {'rock': ['scissors', 'lizard'],
             'paper': ['rock', 'spock'],
             'scissors': ['paper', 'lizard'],
             'lizard': ['spock', 'paper'],
             'spock': ['scissors', 'rock']}

def rplsls_game():
    player_score, computer_score = 0, 0
    weapons = ['rock', 'paper', 'scissors', 'lizard', 'spock']

    while(abs(player_score - computer_score) < 2):

        print "-----------------------------------------------------------"
        print " Please enter your selection: "
        print " 1 (Rock), 2 (Paper), 3 (Scissors), 4 (Lizard) or 5 (Spock)"

        computer_weapon = weapons[randint(0, 4)]
        weapon = raw_input()
        if weapon in '12345': weapon = weapons[int(weapon) - 1]
        if weapon not in weapons:
            print "invalid input"
            continue        
        print "You selected: " + weapon
        print "Computer selected: " + computer_weapon

        if computer_weapon in beaten_by[weapon]:
            print "You won!"
            player_score += 1
        elif weapon in beaten_by[computer_weapon]:
            print "Computer won!"
            computer_score += 1
        else:
            print "Draw!"

        print "Player: " + str(player_score)
        print "Computer: " + str(computer_score)

    if player_score > computer_score:
        print "Congratulations! you won the game"
    else:
        print "Computer won the game..."

rplsls_game()

Answer 6

您也可以使用raw_input instand：

print("    Please enter your selection: 1 (Rock), 2 (Paper), 3 (Scissors), 4 (Lizard) or 5 (Spock): ")
selection = int(input())

try:
  selection = input("Please enter your selection: 1 (Rock), 2 (Paper), 3 (Scissors), 4 (Lizard) or 5 (Spock): ")
except ...

您完全忘记了异常。

如果你的give_winner函数中的词干太大，请使用dictionar或lambda函数。

一种更有效的编写代码的方法

6 个答案: