Java Stream 中多个对象属性的累积和

Question

我有一个按月年String属性排序的对象列表。

我的对象类定义如下：

public class Obj {
    String year;
    Long membercount;
    Long nonmembercount;
    Double memberpayment;
    Double nonmemberpayment;
}

new Obj("9-2015",100,20,10,5),
new Obj("10-2015",220,40,20,55),
new Obj("11-2015",300,60,30,45),
new Obj("12-2015",330,30,50,6),
new Obj("1-2016",100,10,10,4)
)

membercount我想对、nonmembercount、memberpayment和进行累积和nonmemberpayment。

所以我的新List对象如下所示：

new Obj("9-2015",100,20,10,5)
new Obj("10-2015",320,60,30,60)
new Obj("11-2015",620,120,60,105)
new Obj("12-2015",950,150,110,111)
new Obj("1-2016",1050,160,120,115)

我尝试过，Collectors.summingDouble但它给了我所有的总和，而不是累积的。

我真的很感激任何指点。

Answer 1

Stream API 中不直接支持累积操作，尽管可以通过自定义Collector. 但值得注意的是，\xe2\x80\x99s 已经直接支持数组上的此类操作，这可能足以满足您的情况：

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将你的草图延伸Obj到

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public class Obj {\n    String year;\n    Long membercount;\n    Long nonmembercount;\n    Double memberpayment;\n    Double nonmemberpayment;\n\n    public Obj(String year, long membercount, long nonmembercount,\n            double memberpayment, double nonmemberpayment) {\n        this.year = year;\n        this.membercount = membercount;\n        this.nonmembercount = nonmembercount;\n        this.memberpayment = memberpayment;\n        this.nonmemberpayment = nonmemberpayment;\n    }\n\n    @Override\n    public String toString() {\n        return "Obj("+year+", "+membercount+", "+nonmembercount\n            +", "+memberpayment+", "+nonmemberpayment+\')\';\n    }\n}\n

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解决方案可能如下所示：

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// test data\nList<Obj> list=Arrays.asList(\n    new Obj("9-2015", 100, 20, 10, 5),\n    new Obj("10-2015", 220, 40, 20, 55),\n    new Obj("11-2015", 300, 60, 30, 45),\n    new Obj("12-2015", 330, 30, 50, 6),\n    new Obj("1-2016", 100, 10, 10, 4));\n\n// creating an array as need for the operation, it will contain the\n// result afterwards, whereas the source list is not modified\nObj[] array = list.toArray(new Obj[0]);\n\n// the actual operation\nArrays.parallelPrefix(array, (a,b) -> new Obj(b.year,\n    a.membercount + b.membercount,\n    a.nonmembercount + b.nonmembercount,\n    a.memberpayment + b.memberpayment,\n    a.nonmemberpayment + b.nonmemberpayment\n));\n\n// just print the result\nArrays.asList(array).forEach(System.out::println);\n

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最后一行将打印

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public class Obj {\n    String year;\n    Long membercount;\n    Long nonmembercount;\n    Double memberpayment;\n    Double nonmemberpayment;\n\n    public Obj(String year, long membercount, long nonmembercount,\n            double memberpayment, double nonmemberpayment) {\n        this.year = year;\n        this.membercount = membercount;\n        this.nonmembercount = nonmembercount;\n        this.memberpayment = memberpayment;\n        this.nonmemberpayment = nonmemberpayment;\n    }\n\n    @Override\n    public String toString() {\n        return "Obj("+year+", "+membercount+", "+nonmembercount\n            +", "+memberpayment+", "+nonmemberpayment+\')\';\n    }\n}\n

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虽然此操作不太可能从如此少量元素的并行处理中受益，但遗憾的是此操作没有顺序版本。所以你可能会考虑使用普通的循环解决方案来代替\xe2\x80\xa6

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为了完整起见，这里是一个基于流收集器的累积操作解决方案。与 with 一样Arrays.parallelPrefix，更新函数必须是无副作用和关联的，这是返回具有汇总属性的新对象的函数的情况。

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public static <T> Collector<T,?,List<T>> cumulative(BinaryOperator<T> update) {\n    return Collector.of(ArrayList::new,\n        (l,o) -> {\n            if(!l.isEmpty()) o=update.apply(l.get(l.size()-1), o);\n            l.add(o);\n        },\n        (l,m) -> {\n            if(l.isEmpty()) return m;\n            if(!m.isEmpty()) {\n                T a = l.get(l.size()-1);\n                for(T b: m) l.add(update.apply(a, b));\n            }\n            return l;\n        });\n}\n

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将其与上述设置一起使用：

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List<Obj> result = list.stream().collect(cumulative((a,b) -> new Obj(b.year,\n    a.membercount + b.membercount,\n    a.nonmembercount + b.nonmembercount,\n    a.memberpayment + b.memberpayment,\n    a.nonmemberpayment + b.nonmemberpayment\n)));\n

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