Note3 :《集体智慧编程》用户相似度计算

欧几里德距离评价：

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" 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以经过人们一致评价的物品为坐标轴，然后将参与评价的人绘制到图上，并考察他们彼此之间的距离远近。计算出每一轴向上的差值，求平方之后再相加，最后对总和取平方根。

# -*- coding: UTF-8 -*-

#一个涉及影评者及其对几部影片评分情况的字典
critics={'Lisa Rose': {'Lady in the Water': 2.5, 'Snakes on a Plane': 3.5,
 'Just My Luck': 3.0, 'Superman Returns': 3.5, 'You, Me and Dupree': 2.5,
 'The Night Listener': 3.0},
'Gene Seymour': {'Lady in the Water': 3.0, 'Snakes on a Plane': 3.5,
 'Just My Luck': 1.5, 'Superman Returns': 5.0, 'The Night Listener': 3.0,
 'You, Me and Dupree': 3.5},
'Michael Phillips': {'Lady in the Water': 2.5, 'Snakes on a Plane': 3.0,
 'Superman Returns': 3.5, 'The Night Listener': 4.0},
'Claudia Puig': {'Snakes on a Plane': 3.5, 'Just My Luck': 3.0,
 'The Night Listener': 4.5, 'Superman Returns': 4.0,
 'You, Me and Dupree': 2.5},
'Mick LaSalle': {'Lady in the Water': 3.0, 'Snakes on a Plane': 4.0,
 'Just My Luck': 2.0, 'Superman Returns': 3.0, 'The Night Listener': 3.0,
 'You, Me and Dupree': 2.0},
'Jack Matthews': {'Lady in the Water': 3.0, 'Snakes on a Plane': 4.0,
 'The Night Listener': 3.0, 'Superman Returns': 5.0, 'You, Me and Dupree': 3.5},
'Toby': {'Snakes on a Plane':4.5,'You, Me and Dupree':1.0,'Superman Returns':4.0}}

from math import sqrt

#返回一个有关person1与person2的基于距离的相似度评价
def sim_distance (prefs,person1,person2):
    #得到shared_items的列表
    si={}
    for item in prefs[person1]:
        for item in prefs[person2]:
            si[item]=1

    #如果两者没有共同之处，则返回0
    if len(si)==0: return 0

    #计算所有差值的平方和
    sum_of_squares=sum([ pow(prefs[person1][item]-prefs[person2][item],2)
                         for item in prefs[person1] if item in prefs[person2]])

    return 1/(1+sqrt(sum_of_squares))

print(sim_distance(critics,'Lisa Rose','Gene Seymour'))

皮尔逊相关度评价：

Mick Lasalle为《Superman》评了3分，而Gene Seyour则评了5分，所以该影片被定位中图中的(3,5)处。在图中还可以看到一条直线。

皮尔逊相关系数是判断两组数据与某一直线拟合程度的一种度量。

通常情况下：

相关系数0.8-1.0为极强相关

0.6-0.8为强相关

0.4-0.6为中等程度相关

0.2-0.4为弱相关

0.0-0.2为极弱相关或无相关

最佳拟合线：尽可能地靠近图上的所有坐标点。

修正“夸大分值”情况。

皮尔逊积差系数：

　　数学特征：

　　

　　　　　其中，E是数学期望，cov表示协方差。

　　　　　因为μX = E(X)，σX² = E(X²) − E²(X)，同样地，对于Y，可以写成

　　

　　当两个变量的标准差都不为零，相关系数才有定义。从柯西—施瓦茨不等式可知，相关系数不超过1. 当两个变量的线性关系增强时，相关系数趋于1或-1。当一个变量增加而另一变量也增加时，相关系数大于0。当一个变量的增加而另一变量减少时，相关系数小 于0。当两个变量独立时，相关系数为0.但反之并不成立。 这是因为相关系数仅仅反映了两个变量之间是否线性相关。比如说，X是区间［－１，１］上的一个均匀分布的随机变量。Y = X². 那么Y是完全由X确定。因此Y 和X是不独立的。但是相关系数为0。或者说他们是不相关的。当Y 和X服从联合正态分布时，其相互独立和不相关是等价的。

　　假设有两个变量X、Y，那么两变量间的皮尔逊相关系数可通过以下公式计算：

　　公式一：

　　公式二：

　　公式三：

　　公式四：

　　以上列出的四个公式等价，其中E是数学期望，cov表示协方差，N表示变量取值的个数。

利用公式一代码：

# -*- coding: UTF-8 -*-

#一个涉及影评者及其对几部影片评分情况的字典
critics={'Lisa Rose': {'Lady in the Water': 2.5, 'Snakes on a Plane': 3.5,
 'Just My Luck': 3.0, 'Superman Returns': 3.5, 'You, Me and Dupree': 2.5,
 'The Night Listener': 3.0},
'Gene Seymour': {'Lady in the Water': 3.0, 'Snakes on a Plane': 3.5,
 'Just My Luck': 1.5, 'Superman Returns': 5.0, 'The Night Listener': 3.0,
 'You, Me and Dupree': 3.5},
'Michael Phillips': {'Lady in the Water': 2.5, 'Snakes on a Plane': 3.0,
 'Superman Returns': 3.5, 'The Night Listener': 4.0},
'Claudia Puig': {'Snakes on a Plane': 3.5, 'Just My Luck': 3.0,
 'The Night Listener': 4.5, 'Superman Returns': 4.0,
 'You, Me and Dupree': 2.5},
'Mick LaSalle': {'Lady in the Water': 3.0, 'Snakes on a Plane': 4.0,
 'Just My Luck': 2.0, 'Superman Returns': 3.0, 'The Night Listener': 3.0,
 'You, Me and Dupree': 2.0},
'Jack Matthews': {'Lady in the Water': 3.0, 'Snakes on a Plane': 4.0,
 'The Night Listener': 3.0, 'Superman Returns': 5.0, 'You, Me and Dupree': 3.5},
'Toby': {'Snakes on a Plane':4.5,'You, Me and Dupree':1.0,'Superman Returns':4.0}}

from math import sqrt

#返回p1和p2的皮尔逊相关系数
def sim_pearson(prefs,p1,p2):
    #得到双方都曾评价过的物品列表
    si={}
    for item in prefs[p1]:
        if item in prefs[p2]:
            si[item]=1

    #得到列表元素的个数
    n=len(si)

    #如果两人没有共同之处,则返回0
    if n==0: return 0

    #对所有偏好求和
    sum1=sum([prefs[p1][it] for it in si])
    sum2=sum([prefs[p2][it] for it in si])

    #求平方和
    sum1Sq=sum([pow(prefs[p1][it],2) for it in si])
    sum2Sq=sum([pow(prefs[p2][it],2) for it in si])

    #求乘积之和
    pSum=sum([prefs[p1][it]*prefs[p2][it] for it in si])

    #计算皮尔逊评价值
    num=pSum/n-(sum1*sum2)/(n*n)
    den=sqrt((sum1Sq/n-pow(sum1,2)/(n*n))*(sum2Sq/n-pow(sum2,2)/(n*n)))
    if den==0:
        return 0

    r=num/den
    return r

print(sim_pearson(critics,'Lisa Rose','Gene Seymour'))