c# - 标准偏差计算在 C# 中不起作用？

Question

所以我使用这种方法来计算我的数组的标准差，但它似乎没有给我正确的值。

double numbers[] = new double[10];
double sumOfAllItems = 0;

private double total()
{
    for (int i = 0; i < numbers.Length; i++)
    {
        sumOfAllItems += numbers[i];
    }

    return sumOfAllItems;
}

public double mean()
{
    // working
    **sumOfAllItems = 0;**
    return total() / numbers.Length;
}

// numbers are from (1-10) (too lazy to type up all of them.

public double variance()
{
    // each (value - mean) squared
    double summationsTotal = 0; // (numbers[i] - mean() squared

    for (int i = 0; i < numbers.Length; i++)
    {
        summationsTotal += Math.Pow(numbers[i] - mean(), 2);
    }

    return summationsTotal / (numbers.Length - 1);
}

我手动计算了方差，结果是 9.166666。这是用于示例公式。

但是，当我在我的 GUI 上计算它时，它却给了我 866.25。这段代码有什么问题吗？

Answer 1

问题出在这里:

double sumOfAllItems = 0;

private double total()
{
    for (int i = 0; i < numbers.Length; i++)
    {
        sumOfAllItems += numbers[i];
    }

    return sumOfAllItems;
}

应该是:

double sumOfAllItems = 0;

private double total()
{
   sumOfAllItems = 0;
   for (int i = 0; i < numbers.Length; i++)
    {
        sumOfAllItems += numbers[i];
    }

    return sumOfAllItems;
}

缓存均值而不是在方差函数中重新计算它也会更有效 - 它不会改变。像这样的东西:

public double variance()
{
    // each (value - mean) squared
    double dMean - mean();
    double summationsTotal = 0; // (numbers[i] - mean() squared

    for (int i = 0; i < numbers.Length; i++)
    {
        summationsTotal += Math.Pow(numbers[i] - dMmean, 2);
    }

    return summationsTotal / (numbers.Length - 1);
}