有没有办法找到算术平均值＆QUOT;好＆QUOT;比之和（）/ N [英] Is there any way to find arithmetic mean "better" than sum()/N?

问题描述

假设我们有N个数字（整数，浮点数，任何你想要的），并希望找到自己的算术平均值。最简单的方法是通过值的数目来总结所有的价值和鸿沟：

Suppose we have N numbers(integers, floats, whatever you want) and want to find their arithmetic mean. Simplest method is to sum all values and divide by number of values:

def simple_mean(array[N]): # pseudocode
    sum = 0
    for i = 1 to N
       sum += array[i]
    return sum / N

它工作正常，但需要大整数。 如果我们不希望大整数和我们都是精品与舍入误差，N是2的幂，我们可以使用分而治之：（（A + B）/ 2 + （C + D）/ 2）/ 2 =（A + B + C + D）/ 4 ，（（A + B + C + D）/ 4 +（ E + F + G + H）/ 4）/ 2 =（A + B + C + D + E + F + G + H）/ 8 ，等等。

It works fine, but requires big integers. If we don't want big integers and we are fine with rounding errors, and N is the power of two, we can use 'divide-and-conquer' : ((a+b)/2 + (c+d)/2)/2 = (a+b+c+d)/4, ((a+b+c+d)/4 + (e+f+g+h)/4)/2 = (a+b+c+d+e+f+g+h)/8, so on.

def bisection_average(array[N]):
   if N == 1: return array[1]
   return (bisection_average(array[:N/2])+bisection_average(array[N/2:]))/2

其他方法吗？

Any other ways?

PS。 操场懒惰

推荐答案

克努特列出了下面的方法来计算给定浮点（原上页中的计算机编程的艺术第2卷，1998年版;我下面的适应避免特殊套管第一次迭代）：

Knuth lists the following method for calculating mean and standard deviation given floating point (original on p. 232 of Vol 2 of The Art of Computer Programming, 1998 edition; my adaptation below avoids special-casing the first iteration):

double M=0, S=0;

for (int i = 0; i < N; ++i)
{
    double Mprev = M;
    M += (x[i] - M)/(i+1);
    S += (x[i] - M)*(x[i] - Mprev);
}

// mean = M
// std dev = sqrt(S/N) or sqrt(S/N+1)
// depending on whether you want population or sample std dev

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