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Sweet Snippet 之 方差计算

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方差计算的简单实现

在概率统计中,方差用于衡量一组数据的离散程度,相关的计算公式如下(总体方差):

μ = 1 N i = 1 N x i σ 2 = 1 N i = 1 N ( x i μ ) 2 \begin{aligned}&\mu = \frac{1}{N}\sum_{i = 1}^{N}x_i \\&\sigma^2 = \frac{1}{N}\sum_{i = 1}^{N}(x_i - \mu)^2\end{aligned}

其中 μ \mu 为数据的平均值, 而 σ 2 \sigma^2 即是(总体)方差.

相应的实现代码如下:

-- Lua
function average(values, count)
    local sum = 0
    
    for i = 1, count do
        sum = sum + values[i]
    end
    
    return sum / count
end

function variance(values, count)
    local average = average(values, count)
    local variance = 0
    
    for i = 1, count do
        local delta = values[i] - average 
        variance = variance + delta * delta
    end
    
    return variance / count
end

通常我们需要在获取新样本数据时更新方差,简单的方法就是按照上述公式重新计算一遍,我们可以通过计算数据子集方差的方式来模拟这个过程:

-- Lua
function variance_list(values)
    local ret = {}
    
    for i = 1, #values do
        ret[i] = variance(values, i)
    end
    
    return ret
end

更好的一种方式是通过递推来计算数据子集的方差,这需要对方差的计算公式做一些变形:

σ 2 = 1 N i = 1 N ( x i μ ) 2 σ 2 = 1 N i = 1 N ( x i 2 + μ 2 2 x i μ ) σ 2 = 1 N ( i = 1 N x i 2 + i = 1 N μ 2 i = 1 N 2 x i μ ) σ 2 = 1 N ( i = 1 N x i 2 + N μ 2 2 μ i = 1 N x i ) σ 2 = 1 N ( i = 1 N x i 2 + N μ 2 2 N μ 2 ) σ 2 = 1 N ( i = 1 N x i 2 N μ 2 ) σ 2 = 1 N ( i = 1 N x i 2 N ( i = 1 N x i N ) 2 ) σ 2 = 1 N ( i = 1 N x i 2 ( i = 1 N x i ) 2 N ) \begin{aligned}&\sigma^2 = \frac{1}{N}\sum_{i = 1}^{N}(x_i - \mu)^2 \implies \\&\sigma^2 = \frac{1}{N}\sum_{i = 1}^{N}(x_i^2 + \mu^2 - 2x_i\mu) \implies \\&\sigma^2 = \frac{1}{N}(\sum_{i = 1}^{N}x_i^2 + \sum_{i = 1}^{N}\mu^2 - \sum_{i = 1}^{N}2x_i\mu) \implies \\&\sigma^2 = \frac{1}{N}(\sum_{i = 1}^{N}x_i^2 + N\mu^2 - 2\mu\sum_{i = 1}^{N}x_i) \implies \\&\sigma^2 = \frac{1}{N}(\sum_{i = 1}^{N}x_i^2 + N\mu^2 - 2N\mu^2) \implies \\&\sigma^2 = \frac{1}{N}(\sum_{i = 1}^{N}x_i^2 - N\mu^2) \implies \\&\sigma^2 = \frac{1}{N}(\sum_{i = 1}^{N}x_i^2 - N(\frac{\sum_{i=1}^{N}x_i}{N})^2) \implies \\&\sigma^2 = \frac{1}{N}(\sum_{i = 1}^{N}x_i^2 - \frac{(\sum_{i=1}^{N}x_i)^2}{N})\end{aligned}

基于此,我们就可以递推的计算数据子集的方差了,相关的计算复杂度则降低了一个数量级( O ( n 2 ) O ( n ) O(n^2) \implies O(n) ):

-- lua
function variance_list_recurrence(values)
    local ret = {}
    
    local pre_square_sum = 0
    local pre_sum = 0
    
    for i = 1, #values do
        local val = values[i]
        
        pre_square_sum = pre_square_sum + val * val
        pre_sum = pre_sum + val
        
        ret[i] = (pre_square_sum - (pre_sum * pre_sum / i)) / i
    end
    
    return ret
end

转载:https://blog.csdn.net/tkokof1/article/details/104344680
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