java - RxJava 使用优化请求

Question

今天我试图解决一个小挑战:

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

每个办公室公开一项服务以获得收入。该调用需要一定的延迟(网络、数据库访问……)。

显然，您希望尽快获得全局收入。

首先我在 python 中尝试并取得了不错的结果:

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')

在我的电脑上大约需要 500 毫秒，这是理想情况(因为 500 毫秒是最大延迟)

接下来，我尝试在 java 中使用 RxJava:

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，大约需要 1000 毫秒(有 500 个线程的池!!)。

当然，我尝试了不同数量的线程，但结果最差或相似。

我不想比较 Python 和 Java，我只想:

如果我做错了解释

更好的方法？

此外，python async 只使用一个线程，但在 Java 中我没有找到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

很简单。在 python 方面，你在异步模式下等待(不阻塞) 在 Java 方面，您等待阻塞代码，因此有所不同。

正确的java代码应该是:

package com.test;

import io.reactivex.Flowable;
import io.reactivex.Single;
import io.reactivex.schedulers.Schedulers;
import org.reactivestreams.Publisher;

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


public class TestReactive {

    public static 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 Publisher<Integer> compute() {
            return Single.just("")
                    .delay(this.sleepTime, TimeUnit.MILLISECONDS)
                    .map(x-> {
                        System.out.printf("%s finished in %dms on thread %d%n", this.name, this.sleepTime, Thread.currentThread().getId());
                        return this.revenue;
                    }).toFlowable();
        }
    }

    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(2).runOn(Schedulers.io()).flatMap(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());

    }

}

编辑:我将并行设置为 2，但谁在乎呢，您可以放置​​一个线程，因为这不是 CPU 限制问题。