欧几里得距离评价(Euclidean Distance Score)

recommendations.py

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, 'You, Me and Dupree' : 3.5, 'The Night Listener' : 3.0},
    
    '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, 'Superman Returns' : 4.0, 'You, Me and Dupree' : 2.5, 'The Night Listener' : 4.5},
    
    'Mick LaSalle' : {'Lady in the Water' : 3.0, 'Snakes on a Plane' : 4.0, 'Just My Luck' : 2.0, 'Superman Returns' : 3.0, 'You, Me and Dupree' : 2.0, 'The Night Listener' : 3.0},
    
    'Jack Matthews' : {'Lady in the Water' : 3.0, 'Snakes on a Plane' : 4.0, 'Superman Returns' : 3.0, 'You, Me and Dupree' : 3.5, 'The Night Listener' : 3.0},
    
    'Toby' : {'Snakes on a Plane' : 4.5, 'Superman Returns' : 4.0, 'You, Me and Dupree' : 1.0}
}

EDS.py

from recommendations import critics
from math import sqrt


def sim_distance(prefs, person1, person2):
    si = {}

    for item in prefs[person1]:
        if item in prefs[person2]:
            si[item] = 1

    if len(si) == 0: return 0

    sum_of_squares = sum([pow(prefs[person1][item] - prefs[person2][item], 2) for item in si])

    return 1 / (1 + sqrt(sum_of_squares))

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

基本原理就和它的名字一样，就是通过计算一个坐标轴中，两点之间的距离。

设有点P1(X1, Y1)，点P2(X2, Y2)，SQRT代表平方根，POW代表平方

则两点之间的距离D = SQRT( POW(X1-X2) + POW(Y1-Y2) )

在本例中X与Y分别代表两部电影平分，如果距离D越小，则代表两个人的相似度越近。