#构造一份打分数据集
1 users = {"小明": {"中国合伙人": 5.0, "太平轮": 3.0, "荒野猎人": 4.5, "老炮儿": 5.0, "我的少女时代": 3.0, "肖洛特烦恼": 4.5, "火星救援": 5.0},
"小红":{"小时代4": 4.0, "荒野猎人": 3.0, "我的少女时代": 5.0, "肖洛特烦恼": 5.0, "火星救援": 3.0, "后会无期": 3.0},
"小阳": {"小时代4": 2.0, "中国合伙人": 5.0, "我的少女时代": 3.0, "老炮儿": 5.0, "肖洛特烦恼": 4.5, "速度与激情7": 5.0},
"小四": {"小时代4": 5.0, "中国合伙人": 3.0, "我的少女时代": 4.0, "匆匆那年": 4.0, "速度与激情7": 3.5, "火星救援": 3.5, "后会无期": 4.5},
"六爷": {"小时代4": 2.0, "中国合伙人": 4.0, "荒野猎人": 4.5, "老炮儿": 5.0, "我的少女时代": 2.0},
"小李": {"荒野猎人": 5.0, "盗梦空间": 5.0, "我的少女时代": 3.0, "速度与激情7": 5.0, "蚁人": 4.5, "老炮儿": 4.0, "后会无期": 3.5},
"隔壁老王": {"荒野猎人": 5.0, "中国合伙人": 4.0, "我的少女时代": 1.0, "Phoenix": 5.0, "甄嬛传": 4.0, "The Strokes": 5.0},
"邻村小芳": {"小时代4": 4.0, "我的少女时代": 4.5, "匆匆那年": 4.5, "甄嬛传": 2.5, "The Strokes": 3.0}
}
#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):
"""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):
"""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):
"""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):
"""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
#查找最近邻
def computeNearestNeighbor(username, users):
"""在给定username的情况下，计算其他用户和它的距离并排序"""
distances = []
for user in users:
if user != username:
#distance = manhattan_dis(users[user], users[username])
distance = pearson_dis(users[user], users[username])
distances.append((distance, user))
# 根据距离排序，距离越近，排得越靠前
distances.sort()
return distances
#推荐
def recommend(username, users):
"""对指定的user推荐电影"""
# 找到最近邻
nearest = computeNearestNeighbor(username, users)[0][1]
recommendations = []
# 找到最近邻看过，但是我们没看过的电影，计算推荐
neighborRatings = users[nearest]
userRatings = users[username]
for artist in neighborRatings:
if not artist in userRatings:
recommendations.append((artist, neighborRatings[artist]))
results = sorted(recommendations, key=lambda artistTuple: artistTuple[1], reverse = True)
for result in results:
print(result[0], result[1])
#测试一下
recommend('小阳', users)
后会无期 4.5
匆匆那年 4.0
火星救援 3.5
#简单的张量分解进行打分和推荐
#要用到numpy模块
import numpy
#手写矩阵分解
#现在有很多很方便对高维矩阵做分解的package，比如libmf, svdfeature等
def matrix_factorization(R, P, Q, K, steps=5000, alpha=0.0002, beta=0.02):
Q = Q.T
for step in xrange(steps):
for i in xrange(len(R)):
for j in xrange(len(R[i])):
if R[i][j] > 0:
eij = R[i][j] - numpy.dot(P[i,:],Q[:,j])
for k in xrange(K):
P[i][k] = P[i][k] + alpha * (2 * eij * Q[k][j] - beta * P[i][k])
Q[k][j] = Q[k][j] + alpha * (2 * eij * P[i][k] - beta * Q[k][j])
eR = numpy.dot(P,Q)
e = 0
for i in xrange(len(R)):
for j in xrange(len(R[i])):
if R[i][j] > 0:
e = e + pow(R[i][j] - numpy.dot(P[i,:],Q[:,j]), 2)
for k in xrange(K):
e = e + (beta/2) * (pow(P[i][k],2) + pow(Q[k][j],2))
if e < 0.001:
break
return P, Q.T
#读取user数据并用张量分解进行打分
R = [
[5,3,0,1],
[4,0,3,1],
[1,1,0,5],
[1,0,0,4],
[0,1,5,4],
]
R = numpy.array(R)
N = len(R)
M = len(R[0])
K = 2
P = numpy.random.rand(N,K)
Q = numpy.random.rand(M,K)
nP, nQ = matrix_factorization(R, P, Q, K)
nR = numpy.dot(nP, nQ.T)
array([[ 0.38345373, 2.181972 ],
[ 0.32334816, 1.56283276],
[ 1.99170613, 0.16400981],
[ 1.59666903, 0.14124969],
[ 1.64308192, 1.07125805]])
array([[ 0.38946426, 2.29198167],
[ 0.19720283, 1.18916254],
[ 1.71589715, 1.76060186],
[ 2.48314488, 0.03019937]])
array([[ 5.15038133, 2.67033753, 4.49955112, 1.01806534],
[ 3.70791658, 1.92222735, 3.30635845, 0.85011689],
[ 1.15160585, 0.58780442, 3.70631887, 4.95064787],
[ 0.94558722, 0.48283649, 2.98840431, 3.96902618],
[ 3.0952255 , 1.59792036, 4.70541851, 4.11236178]])
array([[5, 3, 0, 1],
[4, 0, 3, 1],
[1, 1, 0, 5],
[1, 0, 0, 4],
[0, 1, 5, 4]])