Java 8-如何使用流收集器对映射列表进行分组和求和

Question

我有一个地图列表，并希望对某些列进行分组和求和。

地图列表：

double min=100;
double max=999;
List<Map<String,Object>> positions = new ArrayList<Map<String,Object>>();
for (int i=0; i<10; i++) {
    Map<String,Object> positionRow = new HashMap<String,Object>();
    positionRow.put("id", i+1);
    if ((i+1)%2 == 0)
        positionRow.put("Mod", "Even");
    else
        positionRow.put("Mod", "Odd");
    Random r = new Random();
    double randomValue = min + (max - min) * r.nextDouble();
    positionRow.put("price", randomValue);
    positionRow.put("bigger", (randomValue > 300)?"Y":"N");
    positions.add(positionRow);
}

Map<String, List<Map<String, Object>>> grouped = positions.stream()
        .collect(
                Collectors.groupingBy(m -> m.get("Mod").toString()
                        + "<|>" + m.get("bigger").toString()));

{Even<|>Y=[{Mod=Even, price=872.729803251601, bigger=Y, id=2}, {Mod=Even, price=353.6589309614915, bigger=Y, id=4}, {Mod=Even, price=981.8179976373482, bigger=Y, id=6}, {Mod=Even, price=942.3538966530067, bigger=Y, id=8}, {Mod=Even, price=919.0174189044218, bigger=Y, id=10}], Odd<|>Y=[{Mod=Odd, price=663.7589137270676, bigger=Y, id=1}, {Mod=Odd, price=894.1766799283644, bigger=Y, id=3}, {Mod=Odd, price=905.8003509608488, bigger=Y, id=5}, {Mod=Odd, price=758.3085768816934, bigger=Y, id=7}, {Mod=Odd, price=531.1035747782346, bigger=Y, id=9}]}

{Even<|>Y={sum_price=4069.578047, sum_id=30}, Odd<|>Y={sum_price=3753.148096,sum_id=25}}

Answer 1

map 不是一个很好的维护对象的类型。这正是具有名称和声明类型的类和字段被发明的原因。收集器所需的代码反映了这一点：

Map<String, Map<String, Object>> grouped = positions.stream()
.collect(Collectors.groupingBy(m -> m.get("Mod")+"<|>"+m.get("bigger"),
    Collector.of(HashMap::new, (m,p)-> {
        m.merge("sum_price", p.get("price"), (a,b)->((Double)a)+((Double)b));
        m.merge("sum_id", p.get("id"), (a,b)->((Integer)a)+((Integer)b));
    }, (m1,m2)-> {
        m1.merge("sum_price", m2.get("sum_price"), (a,b)->((Double)a)+((Double)b));
        m1.merge("sum_id", m2.get("sum_id"), (a,b)->((Integer)a)+((Integer)b));
        return m1;
    })));