Python Higher-Lower游戏跨模块导入列表重初始化未重置问题

两个核心Python机制导致你遇到这个bug：

• 模块导入缓存机制：Python对同一个模块只会执行一次导入加载，后续所有重复的导入语句只会直接返回内存中已经缓存的模块对象，不会重新执行模块代码生成新的初始数据。你在游戏过程中直接修改了导入的data列表，后续重复写from game_data_test import data根本拿不到未修改的初始列表。
• 引用赋值特性：你写的new_items = data没有创建新列表，只是给内存中同一个列表对象绑定了新变量名，后续对new_items调用remove删除元素时，本质就是在修改导入的原始data列表本身。
• 额外隐患：你用递归调用game()实现重启，多次重启后会出现函数栈溢出问题。

1. 只在文件最顶部保留一次from game_data_test import data导入语句，删掉游戏逻辑里所有重复的导入代码，重复导入没有任何实际作用。
2. 每次启动新游戏对局时，基于原始data生成一份完全独立的副本作为当前局的操作数据源，所有删除操作只作用在副本上，永远不修改原始导入的data对象，这样每局开始拿到的都是完整的初始数据。
3. 用外层循环替代递归实现游戏重启，避免栈溢出问题。

#Higher-Lower Game Project
import random
# 仅顶部导入一次原始数据，全程不修改这个对象
from game_data_test import data

def game():
  print("Welcome to the 'Higher-Lower' game!")
  # 外层循环控制游戏重启
  while True:
    # 每局开始生成原始数据的独立副本
    new_items = [entry.copy() for entry in data]
    round_number = 1
    streak = 0
    status = "win"

    def influencer_selection():
      number_influencers = len(new_items)
      influencer_index = random.randint(0, number_influencers-1)
      selected_influencer = new_items[influencer_index]
      new_items.remove(selected_influencer)
      return selected_influencer

    def influencer_info(influencer_code):
      return (f"{influencer_code['name']}, a {influencer_code['description']}, from {influencer_code['country']}.")

    influencer_a = influencer_selection()
    influencer_a_info = influencer_info(influencer_a)

    while status == "win" and len(new_items) > 0:
      print(f"Round {round_number}:\n\n")
      influencer_b = influencer_selection()
      influencer_b_info = influencer_info(influencer_b)
      print(f"Compare A: {influencer_a_info}")
      print(f"\nAgainst B: {influencer_b_info}\n")

      if int(influencer_a['follower_count']) > int(influencer_b['follower_count']):
        correct_answer = "A"
      else:
        correct_answer = "B"

      answer = input("Who has a higher number of Instagram followers? A or B? \n\n")

      if answer == correct_answer:
        print("\nYou're right!")
        round_number += 1
        streak += 1
        print(f"\nYou are at {streak} correct guesses!\n")
        influencer_a = influencer_b
        influencer_a_info = influencer_b_info
      else:
        print("\nOh no, you're wrong!")
        status = "loss"
        print(f"\nYou have lost. You had a streak of {streak} correct guesses. Congratulations!\n")
        break

    if len(new_items) == 0 and status == "win":
      print(f"\nCongratulations, you have completed the game with a streak of {streak} correct guesses!\n")

    again = input("Do you want to play again? 'Y' or 'N'\n\n")
    if again == "Y":
      # 直接进入下一轮循环即可，不需要递归调用
      continue
    else:
      print("\nThank you for playing!")
      break

game()

注：这里用[entry.copy() for entry in data]生成副本，是因为列表内存储的是字典对象，用推导式复制每个字典生成新列表，就能保证当前局的数据源和原始导入的data完全独立，互不影响。如果后续你不需要修改字典内部的键值对，也可以直接用new_items = data.copy()做列表浅拷贝，效果一致。

内容的提问来源于stack exchange，提问作者AlexE22