遍历列表删除元素异常：{1,3}未被移除问题排查

我写了一段代码，想从combi列表里移除所有存在于teamed中的集合，但运行结果不对——{1,3}明明在teamed里却没被移除，而且循环根本没遍历到这个元素。

原代码如下：

import itertools  # 补充原代码遗漏的导入

teamed = [{1, 2}, {3, 4}, {5, 6}, {8, 7}, {9, 10}, {11, 12}, {1, 3}, {2, 4}, {5, 7}, {8, 6}, {9, 13}, {10, 14}]
game_3 = [1, 2, 3, 4, 5, 6, 7, 8, 11, 12, 13, 14]
combi = [set(combo) for combo in itertools.combinations(game_3, 2)]
for c in combi:
    if c in teamed:
        combi.remove(c)
        
print(combi)

运行后结果仍包含{1,3}：

[{1, 3}, {1, 4}, {1, 5}, {1, 6}, {1, 7}, {8, 1}, {1, 11}, {1, 12}, {1, 13}, {1, 14}, {2, 3}, {2, 5}, {2, 6}, {2, 7}, {8, 2}, {2, 11}, {2, 12}, {2, 13}, {2, 14}, {3, 5}, {3, 6}, {3, 7}, {8, 3}, {11, 3}, {3, 12}, {3, 13}, {3, 14}, {4, 5}, {4, 6}, {4, 7}, {8, 4}, {11, 4}, {4, 12}, {4, 13}, {4, 14}, {5, 7}, {8, 5}, {11, 5}, {12, 5}, {13, 5}, {5, 14}, {6, 7}, {11, 6}, {12, 6}, {13, 6}, {6, 14}, {11, 7}, {12, 7}, {13, 7}, {14, 7}, {8, 11}, {8, 12}, {8, 13}, {8, 14}, {11, 13}, {11, 14}, {12, 13}, {12, 14}, {13, 14}]

直接在遍历列表的同时调用remove修改列表，会导致遍历索引混乱。比如移除第i个元素后，后面的元素会往前移位，原本的第i+1个元素变成了第i个，但循环索引会直接跳到i+1，从而跳过了这个元素——这就是{1,3}没被遍历到的核心原因。

方案1：遍历列表副本 #

遍历combi的副本，修改原列表不会影响遍历过程：

import itertools

teamed = [{1, 2}, {3, 4}, {5, 6}, {8, 7}, {9, 10}, {11, 12}, {1, 3}, {2, 4}, {5, 7}, {8, 6}, {9, 13}, {10, 14}]
game_3 = [1, 2, 3, 4, 5, 6, 7, 8, 11, 12, 13, 14]
combi = [set(combo) for combo in itertools.combinations(game_3, 2)]

# 遍历combi的副本，避免修改原列表干扰遍历
for c in list(combi):
    if c in teamed:
        combi.remove(c)
        
print(combi)

方案2：列表推导式过滤（推荐） #

直接生成新列表过滤掉不需要的元素，代码更简洁高效：

import itertools

teamed = [{1, 2}, {3, 4}, {5, 6}, {8, 7}, {9, 10}, {11, 12}, {1, 3}, {2, 4}, {5, 7}, {8, 6}, {9, 13}, {10, 14}]
game_3 = [1, 2, 3, 4, 5, 6, 7, 8, 11, 12, 13, 14]
combi = [set(combo) for combo in itertools.combinations(game_3, 2)]

# 列表推导式直接过滤掉在teamed中的元素
combi = [c for c in combi if c not in teamed]

print(combi)

方案3：优化查找效率（大数据量场景） #

如果teamed元素很多，列表查找c in teamed效率很低（O(n)），可以把teamed转成frozenset的集合（普通集合不可哈希，不能作为集合元素），提升查找速度：

import itertools

# 将teamed转为frozenset的集合，把查找复杂度降到O(1)
teamed = {frozenset(s) for s in [{1, 2}, {3, 4}, {5, 6}, {8, 7}, {9, 10}, {11, 12}, {1, 3}, {2, 4}, {5, 7}, {8, 6}, {9, 13}, {10, 14}]}
game_3 = [1, 2, 3, 4, 5, 6, 7, 8, 11, 12, 13, 14]
combi = [set(combo) for combo in itertools.combinations(game_3, 2)]

# 过滤时将元素转为frozenset判断
combi = [c for c in combi if frozenset(c) not in teamed]

print(combi)

内容的提问来源于stack exchange，提问作者Karim Omar