[Java] 数据分析--统计

二项分布

• 需求：5个四面体筛子，筛子三面绿色，一面红色，模拟1000000次，统计每次试验红色落地筛子个数的分布
• 实现：用循环实现5个筛子和1000000次试验，定义函数numRedDown模拟5个筛子试验结果，redDown模拟单次试验结果

Simulation.java

 1 import java.util.Random;
 2
 3 public class Simulation{
 4     static final Random RANDOM = new Random();
 5     static final int n = 5;
 6     static final int N = 1000000;
 7
 8     public static void main(String[] args){
 9         double[] dist = new double[n+1];
10         for(int i = 0; i < N ; i++){
11             int x = numRedDown(n);
12             ++dist[x];
13         }
14         for(int i = 0; i <=n;i++){
15             System.out.printf("%4d%8.4f%n",i,dist[i]/N);
16         }
17     }
18
19     static boolean redDown(){
20         int m = RANDOM.nextInt(4);
21         return (m==0);
22     }
23
24     static int numRedDown(int n){
25         int numRed = 0;
26         for(int i = 0; i < n; i++){
27             if(redDown()){
28                 ++numRed;
29             }
30         }
31         return numRed;
32     }
33 }

BinomialDistrabutionTester.java

 1 import org.apache.commons.math3.distribution.BinomialDistribution;
 2
 3 public class BinomialDistributionTester {
 4     static final int n = 5;
 5     static final double p = 0.25;
 6
 7     public static void main(String[] args) {
 8         BinomialDistribution bd = new BinomialDistribution(n, p);
 9         for (int x = 0; x <= n; x++) {
10             System.out.printf("%4d%8.4f%n", x, bd.probability(x));
11         }
12         System.out.printf("mean = %6.4f%n", bd.getNumericalMean());
13         double variance = bd.getNumericalVariance();
14         double stdv = Math.sqrt(variance);
15         System.out.printf("standard deviation = %6.4f%n", stdv);
16     }
17 }

0 0.2381
1 0.3954
2 0.2629
3 0.0880
4 0.0145
5 0.0010

协方差

• 需求：生成1000个随机数对(x,y)，并计算x和y的相关系数
• 实现：Apache Commons Math library 中相应方法

 1 import java.util.Random;
 2 import org.apache.commons.math3.stat.correlation.Covariance;
 3 import org.apache.commons.math3.stat.descriptive.moment.Variance;
 4
 5 public class CorrelationExample {
 6     static final Random RANDOM = new Random();
 7     static double[][] data1 = random(1000);
 8     static double[][] data2 = {{1, 2, 3, 4, 5}, {1, 3, 5, 7, 9}};
 9     static double[][] data3 = {{1, 2, 3, 4, 5}, {9, 8, 7, 6, 5}};
10
11     public static void main(String[] args) {
12         System.out.printf("rho1 = %6.3f%n", rho(data1));
13         System.out.printf("rho2 = %6.3f%n", rho(data2));
14         System.out.printf("rho3 = %6.3f%n", rho(data3));
15     }
16
17     static double[][] random(int n) {
18         double[][] a = new double[2][n];
19         for (int i = 0; i < n; i++) {
20             a[0][i] = RANDOM.nextDouble();
21             a[1][i] = RANDOM.nextDouble();
22         }
23         return a;
24     }
25
26     static double rho(double[][] data) {
27         Variance v = new Variance();
28         double varX = v.evaluate(data[0]);
29         double sigX = Math.sqrt(varX);
30         double varY = v.evaluate(data[1]);
31         double sigY = Math.sqrt(varY);
32         Covariance c = new Covariance(data);
33         double sigXY = c.covariance(data[0], data[1]);
34         return sigXY/(sigX*sigY);
35     }
36 }

rho1 = -0.036
rho2 = 1.000
rho3 = -1.000

正态分布

• 需求：模拟均值16，标准差2.82的正态分布
• 实现：Apache Commons Math library 的 NomorDistribution类

 1 import org.apache.commons.math3.distribution.NormalDistribution;
 2
 3 public class NormalDistributionTester {
 4     static int n = 32;
 5     static double p = 0.5;
 6     static double mu = n*p;
 7     static double sigma = Math.sqrt(n*p*(1-p));
 8
 9     public static void main(String[] args) {
10         NormalDistribution nd = new NormalDistribution(mu, sigma);
11
12         double a = 17.5, b = 21.5;
13         double Fa = nd.cumulativeProbability(a);
14         System.out.printf("F(a) = %6.4f%n", Fa);
15         double Fb = nd.cumulativeProbability(b);
16         System.out.printf("F(b) = %6.4f%n", Fb);
17         System.out.printf("F(b) - F(a) = %6.4f%n", Fb - Fa);
18     }
19 }

F(a) = 0.7021
F(b) = 0.9741
F(b) - F(a) = 0.2720