java streams：直截了當地減少

Question

我有一個MetricGroup流，其中：

public class MetricGroup {

    private int uploadedDocs;
    private long uploadedKbs;

    // getters and setters

}

我需要在一個指標中對所有指標進行sumarize。 我的意思是，我需要將所有metric.uploadedDocs添加到sumMetric.uploadedDocs和metric.uploadedKds到sumMetric.uploadedKbs 。

我發現我需要某種reduce

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

有任何想法嗎？

Answer 1

你可以使用reduce這個重載：

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

像這樣：

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

您還可以使用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

為了避免在reduce調用期間創建多個/多個MetricGroup對象，可以分別進行兩次調用以對UploadedDocs和UploadedKbs求和，然后構造一個表示結果的新MetricGroup 。

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

可以說更具可讀性......

Answer 3

只需傳入一個lambda（將操縱現有的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，真正獲得一個新的MetricGroup（不需要操作現有的MetricGroup）

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

與java流一樣，您實際上不必使用它們。 我建議創建一個簡單的簡化輔助方法：

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);
}

如果你不能改變你開始使用流，你仍然可以使用上面的方法，只需傳遞一個對Stream.iterator()方法的引用：

MetricGroup reduced = reduce(stream::iterator);

Answer 5

如果你想使用簡化，我建議將MetricGroup設為值類型，方法是將字段MetricGroup final，添加零，並用組合方法替換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);
    }
}

這樣可以很好地執行流操作：

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