平均增加数量的变量

Question

我必须报告传入数字的平均值，如何在不使用某种数据结构来跟踪所有值的情况下做到这一点，然后通过将它们求和并除以值数来计算平均值？

Answer 1

Keep the current sum and count. Update both on every incoming number.

avg = sum / count.

Answer 2

只需保持累加总和以及已收到多少个数字，这就是计算平均值所需的全部。

Answer 3

If you have the numbers a[1] a[2] ... a[n] and you know their average is avg(n) = (a[1] + ... + a[n]) / n , then when you get another number a[n + 1] you can do:

avg(n + 1) = (avg(n) * n + a[n + 1]) / (n + 1)

Some floating point errors are unavoidable, but you should test this and see if it's good enough.

To avoid overflow, you could do the division first:

avg(n + 1) = (avg(n) / (n + 1)) * n + (a[n + 1] / (n + 1))

Answer 4

If I'm not totally mistaken, one could calculate the avg(n+1) also this way:

avg(n+1) = (a[1]+ ... + a[n+1]) / (n+1) = 
         = (a[1]+ ... + a[n])/(n+1)   +   a[n+1]/(n+1) = 
         = (n(a[1]+ ... + a[n])/n) / (n+1) + a[n+1]/(n+1) =
         = n*avg(n) / (n+1) + a[n+1]/(n+1) = 

         = n/(n+1) * avg(n) + a[n+1]/(n+1)

so multiply the old avg by n/(n+1) and add the new element divided by n+1 . Depending on how high n will get and how big your values are, this could reduce rounding errors...

EDIT: Of course you have to calculate n/(n+1) using floats, otherwise it will always render 0...

Answer 5

you don't need to keep track the total sum, only counter:

class Averager {
   float currentAverage;
   size_t count;
   float addData (float value) {
       this->currentAverage += (value - this->currentAverage) / ++count;
       return this->currentAverage;
   }
}

from-> prevent long running averaging from overflow?