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Rock Paper Scissors程序不起作用（Python）

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Problems: Program does not seem to accept the integers entered. Won't add to win/loss/draw count and does not display computer choice in debug mode

Basics Design of the Program: 编写一个程序，让用户可以在计算机上玩Rock，Paper，Scissors游戏 . 该计划应如下工作 . 显示一个菜单：

得分：0胜0平0负（D）ebug显示计算机的选择（N）ew游戏（Q）uit

如果用户输入“Q”或“q”，程序将结束 . 对于新游戏，“N”或“n”，对于调试模式，“D”或“d”，其他任何东西都将导致显示错误消息 .

  • 游戏开始时，会生成1到3范围内的随机数 . 如果数字是1，那么计算机选择了摇滚 . 如果数字是2，则计算机选择了纸张 . 如果数字是3，那么计算机选择了剪刀 . （Don 't display the computer'的选择，除非我们处于"D" ebug模式 . ）

  • 用户在键盘上输入他选择的“1-rock”，“2-paper”或“3-scissors” .

  • 显示计算机的选择 .

  • 根据以下规则选择获胜者：•如果一个玩家选择摇滚而另一个玩家选择剪刀，则摇滚获胜 . （岩石砸碎剪刀 . ）•如果一个玩家选择剪刀而另一个玩家选择纸张，那么剪刀就会胜利 . （剪刀剪纸 . ）•如果一个玩家选择纸张而另一个玩家选择摇滚，那么纸张就会赢 . （纸包裹摇滚 . ）•如果两个玩家做出相同的选择，那么游戏就是平局 .

  • 您的计划将保持赢，输和抽奖的总数 .

  • 重新显示菜单并重复游戏循环 .

My Program:

import random

def main():

    continuing = "y"

    win = 0
    lose = 0
    draw = 0

    while continuing == "y":
        print("Score:", win,"wins,", draw, "draws,", lose,"losses")
        print("(D)ebug to show computer's choice")
        print("(N)ew game")
        print("(Q)uit")

        choice = input(" ")

        if choice == "n" or choice == "N":
            win, draw, lose = playgame(win, draw, lose)

        elif choice == "d" or choice == "D":
            win, draw, lose = playgame2(win, draw, lose)

        elif choice == "q" or choice == "Q":
            break


def playgame(win, draw, lose):

    computer = random.randint(1,3)
    player = input("Enter 1 for Rock, 2 for Paper, or 3 for Scissors: ")

    if computer == 1 and player == 2:
        Score = "You won"
        win += 1

    elif computer == 1 and player == 3:
        Score = "You lost"
        lose += 1

    elif computer == 2 and player == 1:
        Score = "You lost"
        lose += 1

    elif computer == 2 and player == 3:
        Score = "You won"
        win += 1

    elif computer == 3 and player == 1:
        Score = "You won"
        win += 1

    elif computer == 3 and player == 2:
        Score = "You lost"
        lose += 1

    elif computer == player:
        Score = "Draw"
        draw += 1

    return (win, draw, lose)

def playgame2(win, draw, lose):

    computer = random.randint(1, 3)
    player = input("Enter 1 for Rock, 2 for Paper, or 3 for Scissors: ")

    if computer == 1 and player == 2:
        Score = "You won"
        print("Computer chose rock")
        win += 1

    elif computer == 1 and player == 3:
        Score = "You lost"
        print("Computer chose rock")
        lose += 1

    elif computer == 2 and player == 1:
        Score = "You lost"
        print("Computer chose paper")
        lose += 1

    elif computer == 2 and player == 3:
        Score = "You won"
        print("Computer chose paper")
        win += 1

    elif computer == 3 and player == 1:
        Score = "You won"
        print("Computer chose scissors")
        win += 1

    elif computer == 3 and player == 2:
        Score = "You lost"
        print("Computer chose scissors")
        lose += 1

    elif computer == player:
        Score = "Draw"
        print("Computer chose the same as you")
        draw += 1

    return (win, draw, lose)


main()
python random return-value

2 回答

  • 4

    我不是Pythonista，但是猜测，input returns strings，你'll need to convert to integer before comparing to the computer' s int .

    我也认为你错过了DRYing up你的代码 - 你应该能够有一个 playgame 方法，它需要一个额外的布尔参数 debugmode ，而不是直接调用print，调用一个间接，例如：

    def debugPrint(debugString, debugMode)
     if debugMode
         print(debugString)

    希望这有道理吗？

    回复于
  • 0

    这适用于Python 2.x，但不适用于Python 3.x在Python 3.x中，input（）返回字符串 . 因此，玩家的输入将是“1”或“2”或“3”的形式 . 由于1和“1”不同，程序不会执行playgame（）和playgame2（）中if和elif块中的任何行 . 这是一个Python 3.x示例：

    >>> a = input("Input = ")
Input = 1
>>> print a
SyntaxError: Missing parentheses in call to 'print'
>>> print(a)
1
>>> a
'1'
>>> type(a)
<class 'str'>

    因此，您应该在任何需要整数输入的地方使用i = int（input（“Input =”）） .

    但是，在Python 2.x中，input（）将1作为1本身而不是“1” . 但是，当您想要将字符串键入为inpu时，您还必须提供引号 . 这是一个例子：

    >>> a1 = input("Input = ")
Input = 1
>>> a1
1
>>> type(a1)
<type 'int'>
>>> #I want to input a string now:
>>> a2 = input("Input = ")
Input = string

Traceback (most recent call last):
  File "<pyshell#12>", line 1, in <module>
    a2 = input("Input = ")
  File "<string>", line 1, in <module>
NameError: name 'string' is not defined
>>> a2 = input("Input = ")
Input = "string"
>>> a2
'string'
>>> type(a2)
<type 'str'>
>>> a3 = raw_input("Input = ")
Input = hello
>>> a3
'hello'
>>> type(a3)
<type 'str'>
>>>

    在Python 2.x中，raw_input（）函数将输入作为字符串 .

    回复于

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