列表<映射<字符串、对象>> 到 列表<映射<字符串、对象>>按键分组，带空值

Question

我有一个用例，我需要从一个查询合并多行与相同的键。

输入(List<Map&lt字符串,Object>祝辞):

[
{"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}
]

期望输出(List

[
{"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}
]

我如何用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(在这种情况下，例如:brand)。正如您所看到的，我总是需要值而不是NULL，但是如果没有值，那么我们可以在输出中使用NULL。

并且在同一列中，同一键总是只有一个值。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，只要它很容易转换为JSON格式的字符串。

Answer 1

您可以使用收集器来实现。与mergeFunction进行映射，根据brand对Map<String,Object>进行分组，然后在merge函数中写入逻辑，对属性

的值进行merge。

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();