Pearson Distance

Pearson Distance:

$\rho_{X,Y} = \frac{cov(X,Y)}{\sigma_{X}\sigma_{Y}}$

where:

1. $cov$ is the covariance

2. $\sigma_{X}$ is the standard deviation of $X$

3. $\sigma_{Y}$ is the standard deviation of $Y$

$cov(X,Y) = E\[(X-\mu_{X})(Y-\mu_{Y})\] = \frac{1}{n}\sum_{i=1}^{n}{(x_{i}-\mu_{X})(y_{i}-\mu_{Y})}$

$\sigma_{X}^2=E\[(X-\mu_{X})^2\] = \frac{1}{n}\sum_{i=1}^{n}{(x_{i}-\mu_{X})^2}$

$\sigma_{Y}^2 = E\[(Y-\mu_{Y})^2\] = \frac{1}{n}\sum_{i=1}^{n}{(y_{i}-\mu_{Y})^2}$

$\mu_{X} = \frac{1}{n}\sum{x_{i}}$

$\mu_{Y} = \frac{1}{n}\sum{y_{i}}$

pearson distance:

$\rho_{X,Y} = \frac{\sum_{i=1}^{n}{(x_{i}-\mu_{X})(y_{i}-\mu_{Y})}}{\sqrt{\sum_{i=1}^{n}{(x_{i}-\mu_{X})^2}}\sqrt{\sum_{i=1}^{n}{(y_{i}-\mu_{Y})^2}}}$

$\rho_{X,Y} = \frac{\sum_{i=1}^{n}{x_{i}y_{i}}-n\mu_{X}\mu_{Y}}{\sqrt{\sum_{i=1}^{n}{x_{i}^2}-n\mu_{X}^2}\sqrt{\sum_{i}^{n}{y_{i}^2}-n\mu_{Y}^2}}$

When we consider $x = X-\mu_{X}$ and $y = Y-\mu_{Y}$ ， then pearson distance is the vectorial angle cosine between $x$ and $y$

$\cos\theta = \frac{x \cdot y}{\|x\| \ast \|y\|}$

When the length of $X$ and $Y$ is 2, then $\rho_{X,Y}$ is either 1 or -1.

The corresponding python code is as follow:

from math import sqrt

def pearson(v1,v2):
    sum1 = sum(v1)
    sum2 = sum(v2)
    sum1Sq = sum([pow(v,2) for v in v1])
    sum2Sq = sum([pow(v,2) for v in v2])
    pSum = sum([v1[i]*v2[i] for i in range(len(v1))])
    num = pSum - sum1*sum2/len(v1)
    den = sqrt(sum1Sq - pow(sum1,2)/len(v1))*sqrt(sum2Sq - pow(sum2,2)/len(v2))
    if den == 0: return 1.0
    return num/den

posted @ 2018-06-04 21:07 LilyEvansHogwarts 阅读(455) 评论(0) 编辑收藏举报

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Pearson Distance

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