一种简单的石纸剪刀游戏
一种简单的石纸剪刀游戏
提问于 2019-11-06 13:46:38
我刚刚完成了基础中级Python课程的学习，并想测试自己。这是我的基本岩石纸剪刀游戏。
def play():
import random
player_score = 0
ai_score = 0
while True:
try:
user_choice = int(input("\nRock: 1"
"\nPaper: 2"
"\nScissor: 3"
"\nWrite the number of the object you choose: "))
except NameError:
print("You can only enter number.")
continue
except ValueError:
print("You can only enter number.")
continue
if user_choice not in range(1, 4):
print("Invalid number, you have to write numbers between 1-3")
continue
rps = {1: "Rock", 2: "Paper", 3: "Scissor"}
ai_choice = random.randint(1, 3)
print(f"\nPlayer's choice: {rps[user_choice]} | AI's choice: {rps[ai_choice]}")
if user_choice == ai_choice:
print("Draw!")
elif user_choice == 1:
if ai_choice == 2:
print("AI's paper wrapped Player's rock, AI won this round!")
ai_score += 1
elif ai_choice == 3:
print("Player's rock crushed AI's scissor, Player won this round!")
player_score += 1
elif user_choice == 2:
if ai_choice == 1:
print("Players's paper wrapped AI's rock, Player won this round!")
player_score += 1
elif ai_choice == 3:
print("AI's scissor cut Player's paper, AI won this round!")
ai_score += 1
elif user_choice == 3:
if ai_choice == 1:
print("AI's rock crushed Player's scissor, AI won this round!")
ai_score += 1
elif ai_choice == 2:
print("Player's scissor cut AI's paper, Player won this round!")
player_score += 1
print(f"\nPlayer's score: {player_score} | AI's score: {ai_score}"
f"\nGame ends when player or AI's score reach 3!")
if player_score == 3:
print("Game over, Player won!")
quit()
elif ai_score == 3:
print("Game over, AI won!")
proceed = input("Wanna try again and show the AI your unmatched and great wisdom?"
"\nWrite Y or N >: ").lower()
if proceed == "y":
play()
elif proceed == "n":
quit()
else:
print("Invalid command, exiting program...")
quit()
if __name__ == "__main__":
play()那么，你怎么看？我怎样才能做得更好，我的错误是什么？
回答 1
Code Review用户
发布于 2019-11-06 16:17:04
我想说，在您的代码中有两个主要问题。
第一个是主循环，它非常不清楚，很多if语句都是用continue和quit()填充的，这是spagetthi代码，它使您的代码更难读，所以很难调试，因为这会跳到任何地方。
第二个问题是缺乏面向对象程序设计( OOP )，这可以很好地适应这种情况。游戏通常使用强大的面向对象语言(Unreal使用C++，Unreal使用C#等)，因为组件之间有很强的交互，并且非常有利于抽象和模块化。公平地说，这并不是真正需要的短期游戏，如岩石剪刀，但它阻止你进一步扩大游戏规模。
我建议您进行很少的设计改进：
将代码拆分为函数，第一种可能是游戏循环中应该使用的游戏获胜条件，以避免不必要的无限循环，这与常量条件(如while True: )相同。将此语句替换为while has_game_ended():，改为has_game_ended，如下所示：
def has_game_ended(player_score: int, ia_score: int) -> bool:
if player_score == 3 or ia_score == 3:
print("Game over: {} won!".format("player" if player_score == 3 else "AI"))
return True
return False然后使用一个枚举来列出您的移动以使它们更易于在代码中重用(论文比1更有意义)，并使用操作符重载启用比较，请注意，这是一个特定于游戏规则的棘手的比较。
from enum import Enum
class Move(Enum):
PAPER = 1
ROCK = 2
SCISSORS = 3
def __lt__(self, other):
return (self is Move.ROCK and other is Move.PAPER) or (self is Move.Paper and other is Move.SCISSORS) or (self is Move.SCISSORS and other is Move.ROCK)现在你可以写：
>>> player_choice = Move.PAPER
>>> ia_choice = Move.ROCK
>>> player_choice < ia_choice
True就像
player_score = player_score + int(player_choice > ia_choice)
ia_score = ia_score + int(ia_choice > player_choice)这很好，因为它不再比较两个整数，而是比较两个Move，这与操作符重载有很大的不同。
避免脚本中的quit()，它完全减少了代码流，并且不允许很好的错误处理，相反，它更倾向于循环条件，或者尝试为意外行为定制异常的you语句(例如，当您使用无效命令时，如下面的示例所示)。
def WrongCommand(Exception):
pass
if __name__ == "__main__":
try:
play()
except WrongCommand:
print("Invalid command, exiting program...")页面原文内容由Code Review提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://codereview.stackexchange.com/questions/231955
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