协同过滤用户相似度度量

Posted 2020-07-08 who_a

闵氏距离(Minkowski Distance)

当r=1时，曼哈顿距离(Manhatten)

当r=2时，欧氏距离(Euclidean)

r=无穷大，上确界距离(Supermum Distance)

皮尔逊相关系数(Pearson CORRELATION Coeffcient)，取值[-1,1]，1表示完全相关，-1表示完全不相关

近似计算公式

余弦相似度计算，取值[-1,1]，1表示完全相似，-1表示完全不相似

 

users = {"Angelica": {"Blues Traveler": 3.5, "Broken Bells": 2.0, "Norah Jones": 4.5, "Phoenix": 5.0, "Slightly Stoopid": 1.5, "The Strokes": 2.5, "Vampire Weekend": 2.0},
         "Bill":{"Blues Traveler": 2.0, "Broken Bells": 3.5, "Deadmau5": 4.0, "Phoenix": 2.0, "Slightly Stoopid": 3.5, "Vampire Weekend": 3.0},
         "Chan": {"Blues Traveler": 5.0, "Broken Bells": 1.0, "Deadmau5": 1.0, "Norah Jones": 3.0, "Phoenix": 5, "Slightly Stoopid": 1.0},
         "Dan": {"Blues Traveler": 3.0, "Broken Bells": 4.0, "Deadmau5": 4.5, "Phoenix": 3.0, "Slightly Stoopid": 4.5, "The Strokes": 4.0, "Vampire Weekend": 2.0},
         "Hailey": {"Broken Bells": 4.0, "Deadmau5": 1.0, "Norah Jones": 4.0, "The Strokes": 4.0, "Vampire Weekend": 1.0},
         "Jordyn":  {"Broken Bells": 4.5, "Deadmau5": 4.0, "Norah Jones": 5.0, "Phoenix": 5.0, "Slightly Stoopid": 4.5, "The Strokes": 4.0, "Vampire Weekend": 4.0},
         "Sam": {"Blues Traveler": 5.0, "Broken Bells": 2.0, "Norah Jones": 3.0, "Phoenix": 5.0, "Slightly Stoopid": 4.0, "The Strokes": 5.0},
         "Veronica": {"Blues Traveler": 3.0, "Norah Jones": 5.0, "Phoenix": 4.0, "Slightly Stoopid": 2.5, "The Strokes": 3.0}
        }#{用户：{作品：评分}}
def manhattan(rating1, rating2):#计算曼哈顿距离
    """Computes the Manhattan distance. Both rating1 and rating2 are dictionaries
       of the form {\'The Strokes\': 3.0, \'Slightly Stoopid\': 2.5}"""
    distance = 0
    commonRatings = False 
    for key in rating1:
        if key in rating2:
            distance += abs(rating1[key] - rating2[key])
            commonRatings = True
    if commonRatings:
        return distance
    else:
        return -1
def pearson(rating1, rating2):#计算Pearson相关系数
    sum_xy = 0
    sum_x = 0
    sum_y = 0
    sum_x2 = 0
    sum_y2 = 0
    n = 0
    for key in rating1:
        if key in rating2:
            n += 1
            x = rating1[key]
            y = rating2[key]
            sum_xy += x * y
            sum_x += x
            sum_y += y
            sum_x2 += pow(x, 2)
            sum_y2 += pow(y, 2)
    # now compute denominator
    denominator = sqrt(sum_x2 - pow(sum_x, 2) / n) * sqrt(sum_y2 - pow(sum_y, 2) / n)
    if denominator == 0:
        return 0
    else:
        return (sum_xy - (sum_x * sum_y) / n)/denominator

 

相似度的选择：

当不同用户对不同商品评价标准的范围不一样时，使用皮尔逊相关系数；

当数据稠密，且属性值大小十分重要，使用欧氏或者曼哈顿距离；

当数据稀疏，存在很多零值，考虑余弦相似度。

来自《A Programmer\'s Guide To Data Mining》