LFM--梯度下降法--实现基于模型的协同过滤
0.引入依赖
import numpy as np # 数值计算、矩阵运算、向量运算
import pandas as pd # 数值分析、科学计算
1.数据准备
# 定义评分矩阵 R
R = np.array([[4, 0, 2, 0, 1],
[0, 2, 3, 0, 0],
[1, 0, 2, 4, 0],
[5, 0, 0, 3, 1],
[0, 0, 1, 5, 1],
[0, 3, 2, 4, 1],
])
# R.shape # (6, 5)
# R.shape[0] # 6
# R.shape[1] # 5
# len(R) # 6
# len(R[0]) # 5
2.算法的实现
"""@输入参数：
R：M*N 的评分矩阵
K：隐特征向量维度
max_iter: 最大迭代次数
alpha：步长lamda：正则化系数
@输出：分解之后的 P，Q
P：初始化用户特征矩
阵 M*K
Q：初始化物品特征矩阵 N*K，Q 的转置是 K*N
"""
# 给定超参数K = 5
max_iter = 5000
alpha = 0.0002
lamda = 0.004
# 核心算法def LMF_grad_desc(R, K=2, max_iter=1000, alpha=0.0001, lamda=0.002):
# 定义基本维度参数 M = len(R)
N = len(R[0])
# P、Q 的初始值随机生成
P = np.random.rand(M, K) Q = np.random.rand(N, K)
Q = Q.T
# 开始迭代
for steps in range(max_iter):
# 对所有的用户 u，物品 i 做遍历，然后对对应的特征向量 Pu、Qi 做梯度下降
for u in range(M):
for i in range(N):
# 对于每一个大于 0 的评分，求出预测评分误差 e_ui
if R[u[i] > 0:
e_ui = np.dot(P[u,:], Q[:,i]) - R[u][i]
# 代入公式，按照梯度下降算法更新当前的 Pu、Qi
for k in range(K): P[u][k] = P[u][k] - alpha * (2 * e_ui * Q[k][i] + 2 * lamda * P[u][k])
Q[k][i] = Q[k][i] - alpha * (2 * e_ui * P[u][k] + 2 * lamda * Q[k][i])
# u,i 遍历完成，所有的特征向量更新完成，可以得到 P、Q，可以计算预测评分矩阵
predR = np.dot(P, Q)
# 计算当前损失函数（所有的预测误差平方后求和）
cost = 0
for u in range(M):
for i in range(N): # 对于每一个大于 0 的评分，求出预测评分误差后，将所有的预测误差平方后求和
if R[u[i] > 0:
cost += (np.dot(P[u,:], Q[:,i]) - R[u][i]) ** 2
# 加上正则化项
for k in range(K): cost += lamda * (P[u][k] ** 2 + Q[k][i] ** 2)
if cost < 0.0001:
# 当前损失函数小于给定的值，退出迭代
break
return P, Q.T, cost
3.测试
P, Q, cost = LMF_grad_desc(R, K, max_iter, alpha, lamda)
print(P)
print(Q)
print(cost)
predR = P.dot(Q.T)
print(R)predR
当 K = 2 时，输出结果如下：
[[1.44372596 1.29573962]
[1.82185633 0.0158696 ]
[1.5331521 0.16327061]
[0.31364667 1.9008297 ]
[1.03622742 2.03603634]
[1.34107967 0.93406796]][[ 0.4501051 2.55477489]
[ 1.18869845 1.20910294]
[ 1.54255106 -0.23514326]
[ 2.33556583 1.21026575]
[ 0.43753164 0.34555928]]
1.0432768290554293
[[4 0 2 0 1]
[0 2 3 0 0]
[1 0 2 4 0]
[5 0 0 3 1]
[0 0 1 5 1]
[0 3 2 4 1]]
array([[3.96015147, 3.2828374 , 1.92233657, 4.9401063 , 1.07943065],
[0.86057008, 2.18482578, 2.80657478, 4.27427181, 0.80260368],
[1.10719924, 2.0198665 , 2.32657341, 3.77837848, 0.72722223],
[4.99736596, 2.6711301 , 0.03684871, 3.03305153, 0.79407969],
[5.66802576, 3.69353946, 1.11967348, 4.8843224 , 1.15695354],
[2.98996017, 2.72352365, 1.84904408, 4.2626503 , 0.90954065]])
当 K = 5 时，输出结果如下：
[[ 0.77991893 0.95803701 0.75945903 0.74581653 0.58070622]
[ 1.51777367 0.66949331 0.89818609 0.23566984 0.56583223]
[ 0.03567022 0.58391558 1.42477223 0.87262652 -0.52553017]
[ 1.24101793 0.86257736 0.73772417 0.18181617 0.97014545]
[ 0.58789616 0.53522492 0.48830352 1.80622908 0.81202167]
[ 1.08640318 0.87660384 0.68935314 0.84506882 0.92284071]]
[[ 1.64469428 1.10535565 0.56686066 0.38656745 1.56519511]
[ 0.61680687 0.57188343 0.49729111 0.9623455 0.43969708]
[ 0.99260822 0.6007452 1.14768173 -0.16998497 -0.14094479]
[ 0.47070988 0.85347655 1.43546859 1.8185161 0.29759968]
[ 0.07923314 0.49412497 0.53285806 0.23753882 -0.05146021]]
0.7478305665280703
[[4 0 2 0 1]
[0 2 3 0 0]
[1 0 2 4 0]
[5 0 0 3 1]
[0 0 1 5 1]
[0 3 2 4 1]]
array([[3.9694342 , 2.37968507, 2.01268221, 3.8040546 , 1.08714641],
[4.72218838, 2.2412959 , 2.81976984, 3.17210672, 0.95653992],
[1.02652007, 1.67315396, 1.94711343, 3.99085212, 1.28488146],
[5.0014878 , 2.22716585, 2.42906339, 2.99867943, 0.91091753],
[3.80452512, 3.00679363, 1.04401937, 4.96078887, 0.95850804],
[4.91762916, 2.73324389, 2.1224277 , 4.06049468, 1.03980543]])
显然后者更符合预期。
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