Question

我最近练习了python，发现自己参与了很多关于简单剪刀石头布游戏的if语句，看起来像这样：我的问题是如何使我的代码更高效，更友好地编写和阅读

 while True:

        player_choice = raw_input("\n1-Rock\n2-Paper\n3-Scissors\n{} choose a number:".format(name))
        player_choice = int(player_choice)

        if player_choice == 1:
            player_choice = Choices.rock
        if player_choice == 2:
            player_choice = Choices.paper
        if player_choice == 3:
            player_choice = Choices.scissors


    # Getting the cpu choice.
        cpu_choice = random.randint(1, 3)
        if cpu_choice == 1:
            cpu_choice = Choices.rock
        if cpu_choice == 2:
            cpu_choice = Choices.paper
        if cpu_choice == 3:
            cpu_choice = Choices.scissors


        if player_choice == cpu_choice:
            print"\n Its a Tie!\n!"

        if player_choice == Choices.paper and cpu_choice == Choices.rock:
            print"\n Congratulations!\n{} you won!".format(name)

        if player_choice == Choices.scissors and cpu_choice == Choices.paper:
            print"\n Congratulations!\n{} you won!".format(name)

        if player_choice == Choices.rock and cpu_choice == Choices.scissors:
            print"\n Congratulations!\n{} you won!".format(name)

        if cpu_choice == Choices.scissors and player_choice == Choices.paper:
            print"\n Too bad!\n{} you lost!".format(name)

        if cpu_choice == Choices.paper and player_choice == Choices.rock:
            print"\n Too bad!\n{} you lost!".format(name)

        if cpu_choice == Choices.rock and player_choice == Choices.scissors:
            print"\n Too bad!\n{} you lost!".format(name)*

Answer 1

您的if语句可以替换为字典。例如，可以使用如下字典将整数映射到特定的Choices属性：

choices = {1: Choices.rock, 2: Choices.paper, 3: Choices.scissors}

现在您可以使用

player_choice = choices[player_choice]

和

cpu_choice = random.choice(choices.values())

从封装的角度来看，Choices对象负责处理此映射确实是责任。如果您要使用实际的enum.Enum object（需要Python 3或安装the enum34 backport package），则可以使用：

player_choice = Choices(player_choice)

但是根据您定义Choices的方式，可以给它一个__call__方法，该方法基本上使用上述映射为您提供相同的结果。

接下来，您可以使用字典来确定获奖者：

# if a player picks the key, and the opponent has picked the value,
# then the player wins.
wins_against = {
    Choices.rock: Choices.scissors,
    Choices.paper: Choices.rock,
    Choices.scissors: Choices.paper,
}

然后确定获胜者：

if player_choice == cpu_choice:
    print"\n Its a Tie!\n!"
elif wins_against[player_choice] == cpu_choice:
    print"\n Congratulations!\n{} you won!".format(name)
else:  # not the same, and not a win, so the player lost
    print"\n Too bad!\n{} you lost!".format(name)

但是，该映射也可以是您的Choices枚举对象的一部分；给那些wins_against属性：

if player_choice == cpu_choice:
    print"\n Its a Tie!\n!"
elif player_choice.wins_against == cpu_choice:
    print"\n Congratulations!\n{} you won!".format(name)
else:
    print"\n Too bad!\n{} you lost!".format(name)

如果您要使用enum库，则代码可能变为：

from enum import Enum

class Choices(Enum):
    rock = 1, 'scissors'
    paper = 2, 'rock'
    scissors = 3, 'paper'

    def __new__(cls, value, win_against):
        instance = object.__new__(cls)
        instance._value_ = value
        instance._win_against = win_against
        return instance

    @property
    def win_against(self):
        return type(self)[self._win_against]


 while True:
    options = '\n'.join(['{}-{}'.format(c.value, c.name) for c in choices])
    player_choice = raw_input("\n\n{} choose a number:".format(
        options, name))
    try:
        player_choice = int(player_choice)
        player_choice = Choices(player_choice)
    except ValueError:
        print "Not a valid option, try again"
        continue

    cpu_choice = random.choice(list(Choices))

    if player_choice is cpu_choice:
        print"\n Its a Tie!\n!"
    elif player_choice.wins_against is cpu_choice:
        print"\n Congratulations!\n{} you won!".format(name)
    else:  # not the same, and not a win, so the player lost
        print"\n Too bad!\n{} you lost!".format(name)

Answer 2

您应该使用if / elif / else语句-它只会比较条件语句，直到出现真值为止。如果用户输入其他值

，则可能需要显示一条错误消息

    if player_choice == 1:
        player_choice = Choices.rock
    elif player_choice == 2:
        player_choice = Choices.paper
    elif player_choice == 3:
        player_choice = Choices.scissors
    else:
        print "Invalid choice"

如何在Python中获得更多if语句

2 个答案: