java - 如何使用 Java 流计算两个聚合函数？

Question

我有一个这样的“请求”对象列表(id、数量、价格)

List<Request> requests = Arrays.asList(
        new Request(id++, 20, 59.28),
        new Request(id++, 10, 61.23),
        new Request(id++, 30, 60.67),
        new Request(id++, 25, 60.16),
        new Request(id++, 60, 59.67));

我想在一次迭代中计算两个指标 - sum(amount) 和 sum(amount * price)。我需要他们计算平均价格:sum(amount * price)/sum(amount)。

考虑到我想使用 Java 8 流，我发现的唯一变体是将值映射到 Pair 对象并实现自定义消费者:

static class Aggregate implements Consumer<Pair<Double, Double>> {
    private double count = 0L;
    private double sum = 0L;

    public double average() {
        return count > 0 ? sum/(double) count : 0;
    }

    public void combine(Aggregate other) {
        count += other.count;
        sum += other.sum;
    }

    @Override
    public void accept(Pair<Double, Double> data) {
        this.count += data.getLeft();
        this.sum += data.getLeft() * data.getRight();
    }
}

Double avgPrice = requests.stream()
        .map(e -> Pair.<Double, Double>of(e.getAmount(), e.getPrice()))
        .collect(Aggregate::new, Aggregate::accept, Aggregate::combine)
        .average();

这种方法看起来很困惑——我们必须为每个条目创建额外的 Pair 对象 :(

有人知道更好的解决方案吗？

Answer 1

当然。您将需要自定义聚合，但不需要 Pair:

 static class Aggregate {
   private long count = 0L;
   private double sum = 0L;
   double average() { return sum / count; }
   void merge(Aggregate other) {
     count += other.count;
     sum += other.sum;
   }
   void add(int count, double value) {
     this.count += count;
     this.sum += count * value;
   }
 }
}

requests.stream().collect(
   Aggregate::new,
   (aggr, request) -> aggr.add(request.getCount(), request.getPrice()),
   Aggregate::merge)
 .average();

而且您实际上不需要实现 Consumer。

老实说，多遍解决方案可能几乎一样快，也简单得多......

requests.stream()
      .mapToDouble(request -> request.getCount() * request.getPrice())
      .sum()
   / requests.stream().mapToLong(Request::getCount).sum();