List<Map<String, Object>> 到 List<Map<String, Object>> 按键分组

Question

我有一个用例，我需要使用相同的键合并查询中的多行。

输入（列表

[
{"brand": "A", "sum of A": NULL, "sum of B": NULL, "sum of C": NULL, "sum of D": NULL},
{"brand": "B", "sum of A": 2700, "sum of B": 1500, "sum of C": NULL, "sum of D": NULL},
{"brand": "B", "sum of A": NULL, "sum of B": NULL, "sum of C": 1700, "sum of D": NULL},
{"brand": "C", "sum of A": 3000, "sum of B": NULL, "sum of C": NULL, "sum of D": NULL},
{"brand": "C", "sum of A": NULL, "sum of B": NULL, "sum of C": 6000, "sum of D": NULL},
{"brand": "D", "sum of A": 1000, "sum of B": 2000, "sum of C": 4000, "sum of D": 600}
]

预期输出（列表

[
{"brand": "A", "sum of A": NULL, "sum of B": NULL, "sum of C": NULL, "sum of D": NULL},
{"brand": "B", "sum of A": 2700, "sum of B": 1500, "sum of C": 1700, "sum of D": NULL},
{"brand": "C", "sum of A": 3000, "sum of B": NULL, "sum of C": 6000, "sum of D": NULL},
{"brand": "D", "sum of A": 1000, "sum of B": 2000, "sum of C": 4000, "sum of D": 600}
]

如何使用 Java 流 API 实现这一点？ 我不能使用 POJO，因为数据总是不同的，但我总是知道键的名称是什么（例如：品牌在这种情况下）。 如您所见，我总是需要值而不是 NULL，但如果没有值，那么我们可以在输出中使用 NULL。

而且同一列中的同一键始终只有一个值。所以这是无效的输入：

[
{"brand": "B", "sum of A": 2700, "sum of B": 1500, "sum of C": NULL, "sum of D": NULL},
{"brand": "B", "sum of A": 1400, "sum of B": NULL, "sum of C": 1700, "sum of D": NULL}
]

我不需要输出的确切返回类型，只要它易于转换为 JSON 格式的字符串即可。

Answer 1

您可以使用Collectors.toMap 和mergeFunction 来完成，根据brand 对Map<String,Object> 进行分组，然后在合并函数中编写逻辑以合并属性的值

Collection<Map<String,Object>> result =list.stream()
    .collect(Collectors.toMap(
            map -> map.get("brand"),
            Function.identity(),
            (existing, replacement) -> {
                //logic to merge m1 & m2
                replacement.forEach((key,val)-> {
                    existing.compute(key, (exKey, exVal) -> /*logic to sum*/)
                });
                return existing;
            }))
    .values();