如何在Python中獲得更多if語句

Question

我最近練習了python，發現自己參與了很多關於簡單剪刀石頭布游戲的if語句，看起來像這樣：我的問題是如何使我的代碼更高效，更友好地編寫和閱讀

 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語句可以用字典代替。 例如，可以使用如下字典將整數映射到特定的Choices屬性：

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

現在您可以使用

player_choice = choices[player_choice]

和

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

從封裝的角度來看，處理此映射實際上應該是Choices對象的責任。 如果要使用實際的enum.Enum對象 （需要Python 3或enum34 backport軟件包的安裝），則可以使用：

player_choice = Choices(player_choice)

但是根據您定義Choices ，您可以給它一個__call__方法，該方法基本上使用上述映射為您提供相同的結果。

接下來，您可以使用字典來確定獲獎者：

# 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,
}

然后確定獲勝者：

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，該映射也可以是您的Choices枚舉對象的一部分； 給那些設置一個wins_against屬性：

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)

如果要使用enum庫，則代碼可能變為：

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

您應該使用if / elif / else語句-它只會比較條件語句，直到出現真值為止。 如果用戶輸入其他值，則可能要顯示一條錯誤消息

    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"