2019独角兽企业重金招聘Python工程师标准>>> 
假设有一个很耗时的运算，单台机器已经没法满足需求，这时你可以想到由多台计算机协作完成。具体怎么做呢。
举个很简单的例子，假设这个耗时的运算是从1加到100000，你现在有两台服务器，可以让这两台服务器分别完成从1加到50000，和从50001加到100000，然后本机完成这两个结果之和。
两台服务器分别启动两个akka Server，同时还有一个CalcActor。这个计算actor接收两个参数：Integer start和Integer end，可以从start一直加到end，最后将结果返回给发送者：getSender().tell(result)。
@Log4j
class CalcActor extends UntypedActor {
@Override
void onReceive(Object message) {
log.debug "CalcActor received: ${message}----self:${getSelf()},sender:${getSender()}"
if (message instanceof String) {
String[] args = message.split(",")
int start = Integer.parseInt(args[0])
int end = Integer.parseInt(args[1])
double result = 0d
println("start calc:" + start + " upto " + end)
start.upto(end) {
result += it
}
sleep(5000) //模拟还要额外耗时5秒
println("result:" + result)
getSender().tell(result)
} else {
unhandled(message)
}
}
}
两个服务器分别为：AkkaServerApp serverA = new AkkaServerApp("sc", "10.68.3.122", 8888, "calc") //AkkaSystemName为sc，ip为10.68.3.122，端口为8888，serviceName为calc。
AkkaServerApp serverA = new AkkaServerApp("sp", "10.68.3.124", 8888, "calc")//AkkaSystemName为sp，ip为10.68.3.124，端口为8888，serviceName为calc。主要的代码在客户端：
public static void main(String[] args) throws Exception {
final AkkaServerApp app = new AkkaServerApp("xwc", "127.0.0.1", 6666, "client");//客户端akka配置
ActorRef remoteCalcA1 = app.getSystem().actorOf(new Props(CalcActor.class)..withDeploy(new Deploy(new RemoteScope(new Address("akka", "sc", "10.68.3.122", 8888)))), "clientCalcA1");//将CalcActor发布到远程10.68.3.122上
ActorRef remoteCalcA2 = app.getSystem().actorOf(new Props(CalcActor.class)..withDeploy(new Deploy(new RemoteScope(new Address("akka", "sc", "10.68.3.122", 8888)))), "clientCalcA2");//将CalcActor发布到远程10.68.3.124上
final List<Future<Double>> frs = new ArrayList<Future<Double>>();//异步返回结果Future存放在list中
//tell只请求，是否响应它完全不知道。ask是请求，并明确知道未来会相应。
// remoteCalcA.tell("1,10000", app.getServerActor());
// remoteCalcB.tell("10001,20000", app.getServerActor());
Future f1 = akka.pattern.Patterns.ask(remoteCalcA1, "1,50000", 150000);//让远程122计算从1加到50000，超时时间为150秒
Future f2 = akka.pattern.Patterns.ask(remoteCalcA1, "50001,100000", 150000);//并发地让远程124计算从50001加到100000，超时时间150秒
frs.add(f1);
frs.add(f2);
Future<Iterable<Double>> future = Futures.sequence(frs, app.getSystem().dispatcher());将未来返回的结果转换成Future<Iterable<Double>>
Future<Double> fr = future.map(new Mapper<Iterable<Double>, Double>() {
@Override
public Double apply(Iterable<Double> parameter) {
Double result = 0d;
for (Double s : parameter) {//计算两个服务器返回的结果
result += s;
}
return result;
}
});
fr.onSuccess(new OnSuccess<Double>() {
@Override
public void onSuccess(Double result) {
System.out.println("云计算返回结果-----" + result);
}
});
}
还可以让服务器并发处理：把给从1加到50000的任务分成5个线程并行处理：1..10000,10001..20000,20001..30000,30001..40000,40001..50000，这样能更好地提高效率。
如果按上面的方法仅仅是发布多个remote actor：
ActorRef remoteCalcAn = app.getSystem().actorOf(new Props(CalcActor.class)..withDeploy(new Deploy(new RemoteScope(new Address("akka", "sc", "10.68.3.122", 8888)))), "clientCalcAn");
是没法提高效率的，因为这时的CalcActor是单线程的，它只会先接收1..10000，处理完后再接收10001..20000并处理。。。。。
使其能够并行处理很简单，创建remoteActor时加上withRoute即可：
ActorRef remoteCalcAn = app.getSystem().actorOf(new Props(CalcActor.class).withRouter(new RoundRobinRouter(5)).withDeploy(new Deploy(new RemoteScope(new Address("akka", "sc", "10.68.3.122", 8888)))), "clientCalcAn"); //RoundRobinRouter的参数5可以理解为分配5个线程并行处理
代码跟上面基本相同
public static void main(String[] args) throws Exception {
final AkkaServerApp app = new AkkaServerApp("xwc", "127.0.0.1", 6666, "client");
ActorRef remoteCalcA1 = app.getSystem().actorOf(new Props(CalcActor.class).withRouter(new RoundRobinRouter(4)).withDeploy(new Deploy(new RemoteScope(new Address("akka", "sc", "10.68.3.122", 8888)))), "clientCalcA1");
ActorRef remoteCalcB1 = app.getSystem().actorOf(new Props(CalcActor.class).withRouter(new RoundRobinRouter(4)).withDeploy(new Deploy(new RemoteScope(new Address("akka", "sp", "10.68.3.124", 8888)))), "clientCalcB1");
final List<Future<Double>> frs = new ArrayList<Future<Double>>();
Future f1 = akka.pattern.Patterns.ask(remoteCalcA1, "1,10000", 150000);
Future f2 = akka.pattern.Patterns.ask(remoteCalcA1, "10001,20000", 150000);
Future f3 = akka.pattern.Patterns.ask(remoteCalcA1, "20001,30000", 150000);
Future f4 = akka.pattern.Patterns.ask(remoteCalcA1, "30001,40000", 150000);
Future f5 = akka.pattern.Patterns.ask(remoteCalcB1, "40001,50000", 150000);
Future f6 = akka.pattern.Patterns.ask(remoteCalcB1, "50001,60000", 150000);
Future f7 = akka.pattern.Patterns.ask(remoteCalcB1, "60001,70000", 150000);
Future f8 = akka.pattern.Patterns.ask(remoteCalcB1, "70001,80000", 150000);
frs.add(f1);
frs.add(f2);
frs.add(f3);
frs.add(f4);
frs.add(f5);
frs.add(f6);
frs.add(f7);
frs.add(f8);
Future<Iterable<Double>> future = Futures.sequence(frs, app.getSystem().dispatcher());
Future<Double> fr = future.map(new Mapper<Iterable<Double>, Double>() {
@Override
public Double apply(Iterable<Double> parameter) {
Double result = 0d;
for (Double s : parameter) {
result += s;
}
return result;
}
});
fr.onSuccess(new OnSuccess<Double>() {
@Override
public void onSuccess(Double result) {
System.out.println("云计算返回从1加到80000的结果-----" + result);
}
});
}
http://m.oschina.net/blog/81118