今天我试图解决一个小挑战：
您是一家拥有500个办公室的大公司，您想计算全球收入（每个办公室的收入总和）。
每个办事处都公开提供服务以获取收益。调用需要一定的延迟（网络，数据库访问等）。
显然，您希望全球收入尽快。
首先，我在python中尝试了不错的结果：
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')
在我的计算机上，大约需要500毫秒，这是理想的情况（因为最大延迟是500毫秒）
接下来，我在Java中使用RxJava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());
}
}
在我的计算机上，大约需要1000毫秒（有500个线程的池！！）。
当然，我尝试使用不同数量的线程，但是结果很差或相似。
我不想比较Python和Java，我只想：
如果我犯错了的解释
更好的方法？
此外，python异步仅使用一个线程，但是在Java中，我没有发现如何不使用多线程来产生类似的结果。
也许有人可以帮助我？ ：-）
答案 0 :(得分：1)
这很简单。在python端，您以异步模式等待（不阻止） 在Java方面，您需要等待阻塞代码，因此会有所不同。
java中正确的代码应为：
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：我将并行数设置为2，但请注意，您可以放置一个线程，因为这不是CPU限制问题。
答案 1 :(得分：0)
经过多次尝试（感谢M.T的帮助），终于有了一个不错的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());
}
}
此代码飞速发展！ 1_000_000个办公室需要2秒！