java流：直接减少

Question

我有一个MetricGroup 流，其中：

public class MetricGroup {

    private int uploadedDocs;
    private long uploadedKbs;

    // getters and setters

}

我需要将所有指标汇总到一个指标中。我的意思是，我需要将所有metric.uploadedDocs 添加到sumMetric.uploadedDocs 和metric.uploadedKds 到sumMetric.uploadedKbs。

我发现我需要某种reduce

Stream.of(new MetricGroup(1,100), new MetricGroup(1,200))
    .reduce(????);

有什么想法吗？

Answer 1

你可以使用reduce这个重载：

T reduce(T identity,
     BinaryOperator<T> accumulator)

像这样：

.reduce(new MetricGroup(0, 0),
        (x, y) -> new MetricGroup(
                      x.getUploadedDocs() + y.getUploadedDocs()
                      x.getUploadedKbs() + y.getUploadedKbs()
                  )
        )

您也可以使用collect 方法：

private static MetricGroup combine(MetricGroup x, MetricGroup y) {
    x.setUploadedDocs(x.getUploadedDocs() + y.getUploadedDocs());
    x.setUploadedKbs(x.getUploadedKbs() + y.getUploadedKbs());
    return x;
}

// ....

.collect(() -> new MetricGroup(0, 0),
    YourClass::combine,
    YourClass::combine
)

Answer 2

为避免在reduce 调用期间创建多个/多个MetricGroup 对象，您可以进行两次单独调用以分别对UploadedDocs 和UploadedKbs 求和，然后构造一个新的MetricGroup 表示结果.

int uploadedDocsSum = source.stream().mapToInt(MetricGroup::getUploadedDocs).sum();
long uploadedKbsSum = source.stream().mapToLong(MetricGroup::getUploadedKbs).sum();
MetricGroup result = new MetricGroup(uploadedDocsSum, uploadedKbsSum);

可以说也更具可读性...

Answer 3

只需传入一个 lambda（将操纵现有的 MetricGroup）

Stream.of(new MetricGroup(1, 100), new MetricGroup(1, 200))
    .reduce((a, b) -> {
      a.setUploadedDocs(a.getUploadedDocs() + b.getUploadedDocs());
      a.setUploadedKbs(a.getUploadedKbs() + b.getUploadedKbs());
      return a;
    });

// Optional[F.MetricGroup(uploadedDocs=2, uploadedKbs=300)]

或者，真正获得一个新的 MetricGroup（不操纵现有的）

Stream.of(new MetricGroup(1, 100), new MetricGroup(1, 200))
        .reduce((a, b) -> new MetricGroup(a.getUploadedDocs() + b.getUploadedDocs(), a.getUploadedKbs() + b.getUploadedKbs()));

Answer 4

与 Java 流一样，您实际上不必使用它们。我建议创建一个简单的归约辅助方法：

public static MetricGroup reduce(Iterable<? extends MetricGroup> metrics){
   int uploadedDocs = 0;
   long uploadedKbs = 0L;
   for(MetricGroup metric : metrics){
       uploadedDocs += metric.getUploadedDocs();
       uploadedKbs += metric.getUploadedKbs();
   }
   return new MetricGroup(uploadedDocs, uploadedKbs);
}

如果您无法更改从流开始的情况，您仍然可以使用上述方法，只需传递对 Stream.iterator() 方法的引用即可：

MetricGroup reduced = reduce(stream::iterator);

Answer 5

如果您想使用归约，我建议将 MetricGroup 设为值类型，方法是将字段设为 final，添加零，并用组合方法替换 setter。

public class MetricGroup {
    private final int uploadedDocs;
    private final long uploadedKbs;

    // obvious constructor
    // getters

    public static final ZERO = new MetricGroup(0, 0);

    public MetricGroup add(MetricGroup a, MetricGroup b) {
        return new MetricGroup(a.uploadedDocs + b.upLoadedDocs,
                               a.uploadedKbs + b.uploadedKbs);
    }

    public MetricGroup uploadOneDoc(long kbs) {
        return new MetricGroup(uploadedDocs + 1, uploadedKbs + kbs);
    }
}

这会让你很好地执行流操作：

MetricGroup sum = metricGroups.stream()
                              .reduce(MetricGroup.ZERO, MetricGroup::add);