Python变量遍历优化：世界杯抽奖程序代码简化问询

1. 批量处理种子队洗牌 #

不要单独定义Seed1到Seed8变量，把所有种子分组放进一个二维列表，通过循环一次性完成洗牌，彻底消除重复代码：

• 用seeds列表存储所有种子队分组
• 遍历列表中的每个分组，执行random.shuffle()

2. 简洁专业的结果打印 #

用f-string结合str.join()替代繁琐的字符串拼接，代码更易读、易维护。

完整优化代码 #

import random

# 玩家列表
players = ["Alice", "Bob", "Charlie", "Delilah"]
random.shuffle(players)

# 所有种子队分组，统一存入二维列表
seeds = [
    ["Brazil", "Argentina", "France", "Spain"],
    ["England", "Germany", "Netherlands", "Portugal"],
    ["Belgium", "Denmark", "Uruguay", "Croatia"],
    ["Serbia", "Switzerland", "Senegal", "Mexico"],
    ["USA", "Poland", "Ecuador", "Morocco"],
    ["Wales", "Japan", "Ghana", "Canada"],
    ["Qatar", "South Korea", "Iran", "Cameroon"],
    ["Australia", "Saudi Arabia", "Tunisia", "Costa Rica"]
]

# 批量洗牌每个种子分组
for group in seeds:
    random.shuffle(group)

# 分配并打印结果（用enumerate高效获取索引）
for idx, player in enumerate(players):
    # 从每个种子分组中提取对应位置的球队
    assigned_teams = [group[idx] for group in seeds]
    # 用join拼接球队列表，f-string格式化输出
    print(f"{player} = {', '.join(assigned_teams)}")

你之前代码报错的原因 #

你尝试的for a in range(1,9): random.shuffle(range[a])存在两个问题：

1. range(1,9)生成的是整数序列，不是存储变量的容器，无法通过range[a]获取Seed1这类变量
2. 动态通过数字拼接变量名不是Python的规范写法，把相关数据放进列表/字典才是正确的做法

内容的提问来源于stack exchange，提问作者Dino