JavaSparkPi程序实现原理

在下载下来的spark里，有个样例程序叫做JavaSparkPi，大意是利用Spark的MapReduce函数求圆周率．

 

代码如下：

 

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package org.apache.spark.examples;

import org.apache.spark.SparkConf;
import org.apache.spark.api.java.JavaRDD;
import org.apache.spark.api.java.JavaSparkContext;
import org.apache.spark.api.java.function.Function;
import org.apache.spark.api.java.function.Function2;

import java.util.ArrayList;
import java.util.List;

/** 
 * Computes an approximation to pi
 * Usage: JavaSparkPi [slices]
 */
public final class JavaSparkPi {

  public static void main(String[] args) throws Exception {
    SparkConf sparkConf = new SparkConf().setAppName("JavaSparkPi");
    		sparkConf.setMaster("local");
    JavaSparkContext jsc = new JavaSparkContext(sparkConf);

    int slices = (args.length == 1) ? Integer.parseInt(args[0]) : 2;
    int n = 3000000 * slices;
    List<Integer> l = new ArrayList<Integer>(n);
    for (int i = 0; i < n; i++) {
      l.add(i);
    }

    JavaRDD<Integer> dataSet = jsc.parallelize(l, slices);

    int count = dataSet.map(new Function<Integer, Integer>() {
      @Override
      public Integer call(Integer integer) {
        double x = Math.random() * 2 - 1;
        double y = Math.random() * 2 - 1;
        return (x * x + y * y < 1) ? 1 : 0;
      }
    }).reduce(new Function2<Integer, Integer, Integer>() {
      @Override
      public Integer call(Integer integer, Integer integer2) {
        return integer + integer2;
      }
    });

    System.out.println("Pi is roughly " + 4.0 * count / n);

    jsc.stop();
  }
}

 代码一开始构造了一个很大的集合．然后利用Map函数迭代，并随机采样坐标点．

 

实现背景几何解剖大致如下

取圆心x,y正负１区间为正方形，那么正方形面积为４．

取半径为１圆，圆心坐标为０，０．那么圆形面积为3.141........，也就是元周率．

 

代码开始随机采样坐标点，并判断坐标点是否在圆内．

    double x = Math.random() * 2 - 1;
        double y = Math.random() * 2 - 1;
        return (x * x + y * y < 1) ? 1 : 0;

随机构造X,Y,Math.random只会返回小于１的数，所以后面的乘以２减去１，必然是在正方形内．

x*x+y*y=1反映的是坐标是否在圆周上．那么<1自然就是判断是否在圆内部了．

假设在圆内，就返回１，否则返回０，结合后面的reduce就可以得到总共有多少个点是在圆内的．

 

已知合计n个采样点，共count个在圆内的点．

那么count/n就可以得出　采样点在圆内的合计数　所在　总共采样点个数的比例．利用这个比例去乘以正方形面积．就可以得到元周率近似值．

 

结论，当采样数越大，得出的圆周率越精确．