相似度与距离计算python代码实现

 #定义几种距离计算函数
 #更高效的方式为把得分向量化之后使用scipy中定义的distance方法
 
 from math import sqrt
 def euclidean_dis(rating1, rating2):    #欧式距离计算
     """计算2个打分序列间的欧式距离. 输入的rating1和rating2都是打分dict
        格式为{'小时代4': 1.0, '疯狂动物城': 5.0}"""
     distance = 0
     commonRatings = False
     for key in rating1:
         if key in rating2:
             distance += (rating1[key] - rating2[key])^2
             commonRatings = True
     #两个打分序列之间有公共打分电影
     if commonRatings:
         return distance
     #无公共打分电影
     else:
         return -1
 
 def manhattan_dis(rating1, rating2):    #曼哈顿距离计算
     """计算2个打分序列间的曼哈顿距离. 输入的rating1和rating2都是打分dict
        格式为{'小时代4': 1.0, '疯狂动物城': 5.0}"""
     distance = 0
     commonRatings = False
     for key in rating1:
         if key in rating2:
             distance += abs(rating1[key] - rating2[key])
             commonRatings = True
     #两个打分序列之间有公共打分电影
     if commonRatings:
         return distance
     #无公共打分电影
     else:
         return -1
 
 def cos_dis(rating1, rating2):   #余弦相似度计算
     """计算2个打分序列间的cos距离. 输入的rating1和rating2都是打分dict
        格式为{'小时代4': 1.0, '疯狂动物城': 5.0}"""
     distance = 0
     dot_product_1 = 0
     dot_product_2 = 0
     commonRatings = False
 
     for score in rating1.values():
         dot_product_1 += score^2
     for score in rating2.values():
         dot_product_2 += score^2
 
     for key in rating1:
         if key in rating2:
             distance += rating1[key] * rating2[key]
             commonRatings = True
     #两个打分序列之间有公共打分电影
     if commonRatings:
         return 1-distance/sqrt(dot_product_1*dot_product_2)
     #无公共打分电影
     else:
         return -1
 
 def pearson_dis(rating1, rating2):  #皮尔逊相似度计算
     """计算2个打分序列间的pearson距离. 输入的rating1和rating2都是打分dict
        格式为{'小时代4': 1.0, '疯狂动物城': 5.0}"""
     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