RxJava使用优化请求

Question

今天我试着解决一个小小的挑战:

您是一家拥有500个办事处的大公司,您想要计算全球收入(每个办事处的收入总和).

每个办公室都提供服务以获得收入.该呼叫需要一定的延迟(网络,数据库访问,......).

显然,您希望尽可能快地获得全球收入.

首先我在python中尝试了非常好的结果:

import asyncio
import time

DELAYS = (475, 500, 375, 100, 250, 125, 150, 225, 200, 425, 275, 350, 450, 325, 400, 300, 175)


class Office:

    def __init__(self, delay, name, revenue):
        self.delay = delay
        self.name = name
        self.revenue = revenue

    async def compute(self):
        await asyncio.sleep(self.delay / 1000)
        print(f'{self.name} finished in {self.delay}ms')
        return self.revenue


async def main(offices, totest):
    computed = sum(await asyncio.gather(*[o.compute() for o in offices]))
    verdict = ['nok', 'ok'][computed == totest]
    print(f'Sum of revenues = {computed} {verdict}')


if __name__ == "__main__":
    offices = [Office(DELAYS[i % len(DELAYS)], f'Office-{i}', 3 * i + 10) for i in range(500)]
    totest = sum(o.revenue for o in offices)
    start = time.perf_counter()
    asyncio.run(main(offices, totest))
    end = time.perf_counter()
    print(f'Ends in {(end-start)*1000:.3f}ms')

在我的电脑上需要大约500ms,这是理想的情况(因为500ms是最大延迟)

接下来,我尝试使用RxJava的java:

import java.util.concurrent.TimeUnit;

public class Office {
    private int sleepTime;
    private String name;
    private int revenue;

    public Office(int sleepTime, String name, int revenue) {
        this.sleepTime = sleepTime;
        this.name = name;
        this.revenue = revenue;
    }

    public int getRevenue() {
        return revenue;
    }

    public int compute() {
        try {
            TimeUnit.MILLISECONDS.sleep(this.sleepTime);
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
        System.out.printf("%s finished in %dms on thread %d%n", this.name, this.sleepTime, Thread.currentThread().getId());
        return this.revenue;
    }
}

import io.reactivex.Flowable;
import io.reactivex.schedulers.Schedulers;

import java.time.Duration;
import java.time.Instant;
import java.util.ArrayList;

public class Tester {
    private static int[] DELAYS = {475, 500, 375, 100, 250, 125, 150, 225, 200, 425, 275, 350, 450, 325, 400, 300, 175};

    public static void main(String[] args) {
        final ArrayList<Office> offices = new ArrayList<>();

        for (int i = 0; i < 500; i++) {
            offices.add(new Office(DELAYS[i % DELAYS.length], String.format("Office-%d", i), 3 * i + 10));
        }
        int totest = offices.stream().mapToInt(Office::getRevenue).sum();

        final Instant start = Instant.now();
        final Flowable<Office> officeObservable = Flowable.fromIterable(offices);
        int computation = officeObservable.parallel(500).runOn(Schedulers.io()).map(Office::compute).reduce(Integer::sum).blockingSingle();
        boolean verdict = computation == totest;
        System.out.println("" + computation + " " + (verdict ? "ok" : "nok"));
        final Instant end = Instant.now();

        System.out.printf("Ends in %dms%n", Duration.between(start, end).toMillis());

    }
}

在我的电脑上,它需要大约1000毫秒(有500个线程池!).

当然,我试过不同数量的线程,但结果最差或相似.

我不想比较Python和Java,我只想要:

解释我是否犯过错误

更好的方法？

此外,python async只使用一个线程但在Java中我没有发现如何不使用多线程来获得类似的结果.

也许有人可以帮助我？:-)

Answer 1

经过多次尝试（感谢M.T的帮助），我终于有了一个很好的Java实现！

public class Office {
    private int sleepTime;
    private int revenue;

    public Office(int sleepTime, int revenue) {
        this.sleepTime = sleepTime;
        this.revenue = revenue;
    }

    public int getRevenue() {
        return revenue;
    }

    public Single<Integer> compute() {
        return Single.timer(sleepTime, TimeUnit.MILLISECONDS).map(l -> this.revenue);
    }
}


public class Tester {
    private static int[] DELAYS = {475, 500, 375, 100, 250, 125, 150, 225, 200, 425, 275, 350, 450, 325, 400, 300, 175};

    public static void main(String[] args) {
        final ArrayList<Office> offices = new ArrayList<>();

        for (int i = 0; i < 1_000_000; i++) {
            offices.add(new Office(DELAYS[i % DELAYS.length], 1));
        }
        int totest = offices.stream().mapToInt(Office::getRevenue).sum();

        final Instant start = Instant.now();
        final Flowable<Office> officeObservable = Flowable.fromIterable(offices);
        int computation = officeObservable.flatMapSingle(Office::compute).reduce(Integer::sum).blockingGet();
        boolean verdict = computation == totest;
        System.out.println("" + computation + " " + (verdict ? "ok" : "nok"));
        final Instant end = Instant.now();

        System.out.printf("Ends in %dms%n", Duration.between(start, end).toMillis());
    }
}

这段代码速度非常快！1_000_000 个办事处需要 2 秒！