我有一个像这样的“请求”对象列表(id,金额,价格)

List 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 – 总和(金额)和总和(金额*价格).我需要它们来计算平均价格：总和(金额*价格)/总和(金额).

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对象并实现自定义使用者：

static class Aggregate implements Consumer> {

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 data) {

this.count += data.getLeft();

this.sum += data.getLeft() * data.getRight();

}

}

Double avgPrice = requests.stream()

.map(e -> Pair.of(e.getAmount(), e.getPrice()))

.collect(Aggregate::new, Aggregate::accept, Aggregate::combine)

.average();

这种方法看起来很混乱 – 我们必须为每个条目创建额外的Pair对象:(

有谁知道更好的解决方案？

最佳答案 当然.您需要自定义聚合,但不需要对：

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();

而且您实际上并不需要实现Consumer.

老实说,多通道解决方案可能非常快,也更简单……

requests.stream()

.mapToDouble(request -> request.getCount() * request.getPrice())

.sum()

/ requests.stream().mapToLong(Request::getCount).sum();