Loading... **Update: 在实际应用中这种迭代求取平均值的方法仅用来规避求和溢出的问题，精度上略有损失** ## 平均值 有数列$\{x_i\}$，数列前$ t$个的均值为$M_t$，则 $$M_t=\frac{\sum_{i=1}^{t}x_i}{t}$$ $M_t$还有一种迭代的求取方法， $$M_t=\frac{t-1}{t}M_{t-1}+\frac{1}{t}x_t$$ ## 推导过程 有 $$M_t={(\sum_{i=1}^{t}x_i)}/{t}$$ 将$\sum_{i=1}^{t}x_i$拆成$\sum_{i=1}^{t-1}x_i$和$x_t$的和，即 $$\begin{aligned} &M_t &= & (\sum_{i=1}^{t}x_i)/{t} \\ \\ & &=& (\sum_{i=1}^{t-1}x_i+x_t)/t \\ \\ & &=& \frac{1}{t}\sum_{i=1}^{t-1}x_i+\frac{1}{t}x_t\end{aligned}$$ 由$M_{t-1}=(\sum_{i=1}^{t-1}x_i)/(t-1)$可得$\sum_{i=1}^{t-1}x_i=M_{t-1}(t-1)$，带入上式， $$\begin{aligned} &M_t &= & \frac{1}{t}\sum_{i=1}^{t-1}x_i+\frac{1}{t}x_t \\ \\ & &=& \frac{t-1}{t}M_{t-1}+\frac{1}{t}x_t\end{aligned}$$ ## 代码实现 迭代函数 ```cpp void iterative_mean(float& average, const float num, const int index) { float index_1 = static_cast<float>(index + 1); average = average * static_cast<float>(index) / index_1 + num / index_1; } ``` 示例 ```cpp void iterative_mean(float& average, const float num, const int index) { float index_1 = static_cast<float>(index + 1); average = average * static_cast<float>(index) / index_1 + num / index_1; } int main() { float a[10] = {0}; const int count = 10; for (int i = 0; i < count; ++i) { a[i] = i+1; std::cout << a[i] << " "; } std::cout << std::endl; float iter_mean = 0; float acc_mean = 0; for (int i = 0; i < count; ++i) { // 迭代求平均值 iterative_mean(iter_mean, a[i], i); // 累加 acc_mean += a[i]; } // 累加计算平均值 acc_mean /= count; std::cout << "accumulative mean: " << acc_mean << std::endl; std::cout << "iterative mean: " << iter_mean << std::endl; return 0; } ``` 输出： ``` 1 2 3 4 5 6 7 8 9 10 accumulative mean: 5.5 iterative mean: 5.5 ``` Last modification：August 16th, 2018 at 10:28 am © 允许规范转载
2 comments
推导过程太复杂了。
前t个数的和是Mtt，加上第t+1个数再求平均就得到Mt+1了，Mt+1=(Mtt+xt+1)/(t+1)
是的，谢谢ヾ(≧∇≦*)ゝ