Python实现石头剪刀布游戏：如何优化冗长的if-else语句？

嘿，作为Python新手能写出石头剪刀布游戏已经很赞啦！你提到的超长if-else确实可以用更简洁、易维护的方式替代，下面分享几种实用的优化思路：

方法1：用字典映射胜负规则 #

我们可以把所有胜负关系提前存在字典里，直接通过键值对快速判断结果，代码会清爽很多：

import random

options = ["rock", "paper", "scissors"]
# 定义胜负映射：key是玩家出的选项，value是它能击败的电脑选项
win_conditions = {
    "rock": "scissors",
    "paper": "rock",
    "scissors": "paper"
}

played_hand = input("Play rock, paper or scissors please: ").lower()
computer_hand = random.choice(options)  # 用random.choice替代手动计算索引，更简洁

if played_hand not in options:
    print("Invalid input! Please choose rock, paper or scissors.")
else:
    print(f"The computer played {computer_hand}.")
    if played_hand == computer_hand:
        print("It's a tie!")
    elif win_conditions[played_hand] == computer_hand:
        print("You win!")
    else:
        print("You lose!")

这种方式的好处是规则清晰易维护，以后要改规则或者扩展选项（比如加lizard、spock），只需要修改字典就行，不用堆一堆if-else。

方法2：利用索引计算胜负 #

石头剪刀布的胜负是循环克制的，我们可以通过选项的索引来计算：

• 玩家索引 - 电脑索引 的结果如果是1或者-2，玩家赢
• 如果是0，平局
• 其他情况玩家输

代码示例：

import random

options = ["rock", "paper", "scissors"]
played_hand = input("Play rock, paper or scissors please: ").lower()

if played_hand not in options:
    print("Invalid input! Please choose rock, paper or scissors.")
else:
    player_idx = options.index(played_hand)
    computer_idx = random.randint(0, 2)
    computer_hand = options[computer_idx]
    
    print(f"The computer played {computer_hand}.")
    if player_idx == computer_idx:
        print("It's a tie!")
    elif (player_idx - computer_idx) % 3 == 1:
        print("You win!")
    else:
        print("You lose!")

这种方式利用了数学规律，代码更简洁，不需要额外维护规则字典，特别适合这种有循环克制关系的游戏。

方法3：使用枚举类（进阶版） #

如果想让代码更具可读性和规范性，可以用Python的enum模块定义枚举类，把每个选项和对应的克制关系绑定：

import random
from enum import Enum

class Hand(Enum):
    ROCK = ("rock", "scissors")
    PAPER = ("paper", "rock")
    SCISSORS = ("scissors", "paper")
    
    def __init__(self, name, beats):
        self.name = name
        self.beats = beats

# 把枚举转换为方便查找的字典
hand_map = {hand.name: hand for hand in Hand}

played_hand_name = input("Play rock, paper or scissors please: ").lower()

if played_hand_name not in hand_map:
    print("Invalid input! Please choose rock, paper or scissors.")
else:
    player_hand = hand_map[played_hand_name]
    computer_hand = random.choice(list(Hand))
    
    print(f"The computer played {computer_hand.name}.")
    if player_hand == computer_hand:
        print("It's a tie!")
    elif computer_hand.name == player_hand.beats:
        print("You win!")
    else:
        print("You lose!")

这种方式适合想要代码更严谨、可扩展的场景，枚举类让每个选项的职责更清晰。

不管哪种方法，都比一堆嵌套的if-else要好维护得多，你可以根据自己的需求选择合适的方式～

内容的提问来源于stack exchange，提问作者mark olieman