比较两个字典列表，如果key和value存在，则从列表中删除一个字典

def remove_matching_dicts(list1, list2, key):
    # 创建list2中所有键值对的集合
    set2 = {(d[key], d) for d in list2}
    
    # 过滤list1，只保留那些键值对不在set2中的字典
    list1 = [d for d in list1 if (d[key], d) not in set2]
    
    return list1

# 示例数据
list1 = [{'id': 1, 'name': 'Alice'}, {'id': 2, 'name': 'Bob'}, {'id': 3, 'name': 'Charlie'}]
list2 = [{'id': 2, 'name': 'Bob'}, {'id': 4, 'name': 'David'}]

# 执行函数
result = remove_matching_dicts(list1, list2, 'id')
print(result)  # 输出: [{'id': 1, 'name': 'Alice'}, {'id': 3, 'name': 'Charlie'}]

def remove_matching_dicts_safely(list1, list2, key):
    set2 = {(d.get(key), d) for d in list2 if key in d}
    list1 = [d for d in list1 if (d.get(key), d) not in set2]
    return list1