java - 在 Java 中，如何在不使用数组的情况下获得标准差？

Question

使用双 float 据计算任何给定输入的计数、平均值和标准差。

每次运行该程序时，它都会给出我的计数和平均值，但我的标准差显示为 NaN。

import java.util.Scanner;

public class Final 
{
    public static void main (String[]args)
    {
        Scanner in = new Scanner(System.in);

        System.out.print("Enter a range of digits to recieive 1. Count , 2. Average , 3. StdDvn");

        double input = in.nextDouble();

        double count = 1;
        double sum = 0;
        double sumsquared = sum * sum;
        double std = 0;

        while (in.hasNextDouble())
        { double value = in.nextDouble();

            sum = input += value;
            count++;
        }

        double average = sum / count;
        std = Math.sqrt((sumsquared-(sum/count)) / (count - 1));

        System.out.println("Your count of numbers is : " + count);
        System.out.println("Your average of numbers is : " + average);
        System.out.println("Your standard deviation of numbers is : " + std);
    }
}

Answer 1

您的 sumsquared 变量始终为 0，因为您在初始化 double sum=0 后立即从 0*0 计算它;。

这部分应移至求和下方。

要使用循环计算没有数组的标准差，您需要知道以下 3 个值:

• 输入了多少个数字。
• 所有数字的总和。
• 数字的平方和。

公式:E[X^2]-(E[X])^2 参见 wikipedia 。^2 当然意味着平方。

public static void main (String[]args)
{
    Scanner in = new Scanner(System.in);

    double count = 10.0;   // how many numbers are entered, e.g.: 10
    double sum1 = 0.0;    // sum of the numbers
    double sum2 = 0.0;    // sum of the squares
    System.out.println("Enter 10 numbers: ");
    for (int i=0; i < 10; i++) {
      double n = in.nextDouble();
      sum1 += n;
      sum2 += n * n;
    }
    double average = sum1 / count;
    double variance = (count * sum2 - sum1 * sum1) / (count * count);
    double stddev = Math.sqrt(variance);

    System.out.println("Your count of numbers is : " + count);
    System.out.println("Your average of numbers is : " + average);
    System.out.println("Your standard deviation of numbers is : " + stddev);
}