石头剪子布程序无法工作（Python）

Question

问题： 程序似乎无法接受输入的整数。无法增加胜利/平局/失败的计数，也不在调试模式下显示电脑的选择。

程序的基本设计： 编写一个程序，让用户可以和电脑玩石头、剪刀、布的游戏。 程序应该这样运行： 会显示一个菜单：

得分：0 胜利，0 平局，0 失败 (D)ebug 显示电脑的选择 (N)ew game 新游戏 (Q)uit 退出

如果用户输入“Q”或“q”，程序将结束。如果输入“N”或“n”，则开始新游戏；输入“D”或“d”进入调试模式，其他输入会显示错误信息。

1. 当游戏开始时，程序会生成一个1到3之间的随机数字。如果数字是1，电脑选择的是石头；如果是2，电脑选择的是布；如果是3，电脑选择的是剪刀。（除非在“D”ebug模式下，否则不显示电脑的选择。）
2. 用户在键盘上输入自己的选择：“1-石头”、“2-布”或“3-剪刀”。
3. 显示电脑的选择。
4. 根据以下规则选出胜者： • 如果一个玩家选择石头，另一个选择剪刀，那么石头赢。（石头砸碎剪刀。） • 如果一个玩家选择剪刀，另一个选择布，那么剪刀赢。（剪刀剪断布。） • 如果一个玩家选择布，另一个选择石头，那么布赢。（布包裹石头。） • 如果两个玩家选择相同，则游戏平局。
5. 你的程序会持续记录胜利、失败和平局的总数。
6. 重新显示菜单，重复游戏循环。

我的程序：

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()           

Answer 1

在Python 2.x中，这段代码可以正常工作，但在Python 3.x中就不行了。

在Python 3.x中，input()函数返回的是字符串。所以，玩家输入的内容会是“1”或者“2”或者“3”。因为数字1和字符串“1”是不同的类型，所以程序不会执行playgame()和playgame2()中的if和elif代码块。

下面是一个Python 3.x的例子：

>>> 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，任何时候你想要输入一个整数时，都应该使用i = int(input("Input = "))。

但是，在Python 2.x中，input()会把1当作数字1，而不是字符串“1”。不过，如果你想输入一个字符串，就必须加上引号。下面是一个例子：

>>> 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()函数会把输入当作字符串处理。

Answer 2

我不是Python高手，但我猜想，输入的内容会是字符串，所以在和电脑的整数比较之前，你需要把它转换成整数。

我觉得你在代码上有个小窍门没用到，可以让代码更简洁。你应该能写一个单独的playgame方法，增加一个布尔参数debugmode，这样就可以通过这个参数来决定是否直接打印，而不是每次都调用打印。

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

希望这样说你能明白？