Python对比两个字典列表 提取List2中独有字典元素

核心逻辑是先提取List1中所有元素的唯一匹配标识存入集合，再遍历List2筛选出标识不存在于集合中的元素即可，匹配标识由两部分组成：

• 字典中zrepcode键对应的值
• 字典中除zrepcode外的纯数字字符串格式的id键

实现代码 #

List1 = [{"3":[{"period":"P13","value":10,"year":2022}],"zrepcode":"55"},{"1":[{"period":"P10","value":5,"year":2023}],"zrepcode":"55"}]
List2 = [{"1":[{"period":"P1","value":10,"year":2023},{"period":"P2","value":5,"year":2023}],"zrepcode":"55"},{"3":[{"period":"P1","value":4,"year":2023},{"period":"P2","value":7,"year":2023}],"zrepcode":"55"},{"4":[{"period":"P1","value":10,"year":2023}],"zrepcode":"55"}]

# 存储List1所有元素的匹配标识
list1_match_keys = set()
for item in List1:
    zrep_val = item["zrepcode"]
    # 提取字典中的数字格式id键
    num_id_key = [k for k in item.keys() if k != "zrepcode" and k.isdigit()][0]
    list1_match_keys.add((zrep_val, num_id_key))

# 筛选List2独有的元素
res = []
for item in List2:
    zrep_val = item["zrepcode"]
    num_id_key = [k for k in item.keys() if k != "zrepcode" and k.isdigit()][0]
    if (zrep_val, num_id_key) not in list1_match_keys:
        res.append(item)

运行结果 #

执行代码后得到的res和预期结果完全一致：

[{"4":[{"period":"P1","value":10,"year":2023}],"zrepcode":"55"}]

说明：如果后续单个字典中存在多个数字字符串格式的id键，只需要调整id键的提取逻辑，将所有数字键纳入匹配标识的生成规则即可，当前代码适配你给出的单数字键字典结构。

内容的提问来源于stack exchange，提问作者Storyfoyo