# A dictionary of movie critics and their ratings of a small
# set of movies
#字典存储每个人,对电影的打分
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
# Returns a distance-based similarity score for person1 and person2
#欧几里得算法,算出某两个人对电影偏好的相似度
#传入打分表、要计算相似度的两个人
def sim_distance(prefs,person1,person2):
# Get the list of shared_items
si={}
for item in prefs[person1]:
if item in prefs[person2]: si[item]=1
# if they have no ratings in common, return 0
if len(si)==0: return 0
# Add up the squares of all the differences
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+sum_of_squares)
打印相似度
print(sim_distance(critics,'Toby','Jack Matthews'))