如何获得平均值方法以在Java中给出正确的输出

Question

我期望平均方法的结果为3.5，但我得到3.0。 不知道为什么。 我希望Double可以给我结果，但是没有。 java.util.ArrayList;

public class Hangman {
  public static void main(String[] args) {

    ArrayList<Integer> intList = new ArrayList<Integer>();

    intList.add(3);
    intList.add(2);
    intList.add(7);
    intList.add(2);

    System.out.println("The variance number is : ");
    System.out.println(sum(intList));
    System.out.println(intList.size());
    System.out.println(average(intList));
  }

  public static int sum(ArrayList<Integer> intList) {
    int sum = 0;
    for (int counter = 0; counter < intList.size(); counter++) {
        sum = sum + intList.get(counter);
    }
    return sum;
  }

  public static double average(ArrayList<Integer> intList) {
    double avg = sum(intList) / (intList.size());

    return avg;
  }

  public static ArrayList<Double> subtract(ArrayList<Integer> intList) {
    ArrayList<Double> subtracted = new ArrayList<Double>();
    for (double subtract : intList) {
        subtracted.add((double) (subtract - average(intList)));
    }
    return subtracted;
  }

  public static double variance(ArrayList<Integer> intList) {
    double sumDiffsSquared = 0.0;
    double avg = average(intList);
    for (int value : intList) {
        double diff = value - avg;
        diff *= diff;
        sumDiffsSquared += diff;
    }
    return (sumDiffsSquared / (intList.size() - 1));
  }

}

Answer 1

sum需要返回一个double，否则当您执行

sum(intList) / (intList.size());

在您的average方法中，它会将计算得出的值截断为整数，然后将该值放入双精度形式。

Answer 2

您的函数需要返回一个双精度值。

将方法更改为

public static double sum(ArrayList<Integer> intList) {
    double sum = 0;
    for (int counter = 0; counter < intList.size(); counter++) {
        sum = sum + intList.get(counter);
    }
    return sum;
  }

而且你应该还好

Answer 3

尽管avg的类型为double，但sum()方法返回的是Integer值。 所以你从sum()方法返回了一个double

Answer 4

double avg =（double）sum（intList）/（intList.size（））;