在Java流中添加多个字段(以及条件流操作)

Question

假设我有这个课程:

public class Thing {
    private BigDecimal field1;
    private BigDecimal field2;

    private BigDecimal otherField1;
    private BigDecimal otherField2;
    private BigDecimal otherField3;
}

并且在另一个类I for--over a List<Thing> things,将field1和field2添加到迭代完成时返回的总和.

但我想要的是用Java流来实现这一目标.以下是我所拥有的 - 它的工作原理,但我觉得必须有一种方法可以将其浓缩为一个流:

public BigDecimal addFields(List<Thing> things) {
    BigDecimal field1sum = things.parallelStream()
                                     .filter(thing -> thing.getField1() != null)
                                     .map(Thing::getField1)
                                     .reduce(BigDecimal.ZERO, BigDecimal::add);

     BigDecimal field2sum = things.parallelStream()
                                     .filter(thing -> thing.getField2() != null)
                                     .map(Thing::getField2)
                                     .reduce(BigDecimal.ZERO, BigDecimal::add);
     return field1sum.add(field2sum);
}

我怀疑答案是reduce()采用三个参数的方法,其中一个是BiFunction,但我无法弄清楚如何使其工作.编辑:我想我会在传递(x,y) -> x.add(y)到reduce(),但接下来的问题是如何做我map()这两个领域？

另外,是否有可能/如何将此命令式代码转换为功能流？

public BigDecimal addOtherFields(List<Thing> things) {
    BigDecimal result = BigDecimal.ZERO;
    for (Thing thing : things) {
        if (thing.getOtherField2() != null) {
            BigDecimal otherField2 = thing.getOtherField2();
            otherField2 = thing.getOtherField1().subtract(otherField2);
            result = result.add(otherField2);
         } else if (thing.getOtherField3() != null) {
            BigDecimal otherField3 = thing.getOtherField3();
            otherField3 = thing.getOtherField1.subtract(otherField3);
            result = result.add(otherField3);
         }
     }
     return result;
 }

或者,为了更精确一点,我将如何处理基于流的方法中的条件检查？我试图解决filter()问题而没有成功.

Answer 1

使用collect(),使用自定义收藏家帮手,没有什么不同IntSummaryStatistics.

things.stream()
      .collect(ThingCollectorHelper::new, 
               ThingCollectorHelper::accept,
               ThingCollectorHelper::combine);

你的助手类将是这样的:

class ThingCollectorHelper {
    BigDecimal sum1 = BigDecimal.ZERO;
    BigDecimal sum2 = BigDecimal.ZERO;

    void accept(Thing t) {
        if (t.field1 != null)
            sum1 = sum1.plus(t.field1);
        if (t.field2 != null)
            sum2 = sum2.plus(t.field2);
    }

    void combine(ThingCollectorHelper other) {
        sum1 = sum1.plus(other.sum1);
        sum2 = sum2.plus(other.sum2);
    }

}

Answer 2

由于您要统一处理这些字段，您可以考虑flatMap：

public BigDecimal addFields(List<Thing> things) {
    return things.parallelStream()
        .flatMap(thing -> Stream.of(thing.getField1(), thing.getField2()))
        .filter(Objects::nonNull)
        .reduce(BigDecimal.ZERO, BigDecimal::add);
}
public BigDecimal addOtherFields(List<Thing> things) {
    return things.parallelStream()
        .flatMap(thing ->
            Stream.of(thing.getOtherField2(), thing.getOtherField3())
                .filter(Objects::nonNull)
                .map(thing.getOtherField1()::subtract)
        ).reduce(BigDecimal.ZERO, BigDecimal::add);
}