RxJava使用优化请求

Question

今天我试图解决一个小挑战：

你是一家拥有500个办公室的大公司，你想计算全球收入(每个办公室的收入总和)。

每个办公室都公开一项服务以获得收入。呼叫需要一定的延迟(网络、数据库访问…)。

显然，您希望尽快实现全球收入。

首先，我在python中进行了尝试，结果非常好：

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，它大约需要500毫秒，这是理想的情况(因为500毫秒是最大延迟)

接下来，我用RxJava:在java中进行了尝试

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，它大约需要1000ms(有500个线程池！！)。

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：

解释如果我犯了错误

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。

也许有人能帮我？：-)

Answer 1

非常简单。在python端，您在异步模式下等待(不阻塞)在java方面，您使用阻塞代码进行等待，因此有区别。

java中正确的代码应该是：

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：我设置了2的并行，但谁在乎呢，你可以放一个单独的线程，因为这不是CPU限制的问题。

Answer 2

经过多次尝试(感谢M.T的帮助)，我终于有了一个好的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());
}
}

这个代码太快了！1_000_000办公室2个！