近期在看CF的相关论文，《Collaborative Filtering for Implicit Feedback Datasets》思想非常好，非常easy理解。可是从目标函数

是怎样推导出Xu和Yi的更新公式的推导过程却没有非常好的描写叙述。所以以下写一下
推导：
首先对Xu求导：

当中Y是item矩阵，n*f维，每一行是一个item_vec,C^u是n*n维的对角矩阵。
对角线上的每个元素是c_ui,P(u)是n*1的列向量，它的第i个元素为p_ui。
然后令导数=0,可得：

因为x_u和y_i在目标函数中是对称的。所以非常easy得到：

当中X是user矩阵，m*f维度，每一行是一个user_vec，C^i是m*m的对角矩阵。对角线上的每个元素是c_ui。P(i)是m*1的列向量。它的第u和元素是p_ui
然后令导数=0,可得：

以下是论文算法思想的Python实现：

import numpy as np
import scipy.sparse as sparse
from scipy.sparse.linalg import spsolve
import timedef load_matrix(filename, num_users, num_items):t0 = time.time()counts = np.zeros((num_users, num_items))total = 0.0num_zeros = num_users * num_items'''假设要对一个列表或者数组既要遍历索引又要遍历元素时。能够用enumerate，当传入參数为文件时，索引为行号，元素相应的一行内容'''for i, line in enumerate(open(filename, 'r')): #strip()去除最前面和最后面的空格user, item, count = line.strip().split('\t')user = int(user)item = int(item)count = float(count)if user >= num_users:continueif item >= num_items:continueif count != 0:counts[user, item] = counttotal += countnum_zeros -= 1if i % 100000 == 0:print 'loaded %i counts...' % i#数据导入完成后计算稀疏矩阵中零元素个数和非零元素个数的比例，记为alphaalpha = num_zeros / totalprint 'alpha %.2f' % alphacounts *= alpha#用CompressedSparse Row Format将稀疏矩阵压缩counts = sparse.csr_matrix(counts)t1 = time.time()print 'Finished loading matrix in %f seconds' % (t1 - t0)return countsclass ImplicitMF():def __init__(self, counts, num_factors=40, num_iterations=30,reg_param=0.8):self.counts = countsself.num_users = counts.shape[0]self.num_items = counts.shape[1]self.num_factors = num_factorsself.num_iterations = num_iterationsself.reg_param = reg_paramdef train_model(self):#创建user_vectors和item_vectors，他们的元素~N(0,1)的正态分布self.user_vectors = np.random.normal(size=(self.num_users,self.num_factors))self.item_vectors = np.random.normal(size=(self.num_items,self.num_factors))'''要生成非常大的数字序列的时候，用xrange会比range性能优非常多，因为不须要一上来就开辟一块非常大的内存空间，这两个基本上都是在循环的时候用'''for i in xrange(self.num_iterations):t0 = time.time()print 'Solving for user vectors...'self.user_vectors = self.iteration(True, sparse.csr_matrix(self.item_vectors))print 'Solving for item vectors...'self.item_vectors = self.iteration(False, sparse.csr_matrix(self.user_vectors))t1 = time.time()print 'iteration %i finished in %f seconds' % (i + 1, t1 - t0)def iteration(self, user, fixed_vecs):#相当于C的三木运算符。if user=True num_solve = num_users,反之为num_itemsnum_solve = self.num_users if user else self.num_itemsnum_fixed = fixed_vecs.shape[0]YTY = fixed_vecs.T.dot(fixed_vecs)eye = sparse.eye(num_fixed)lambda_eye = self.reg_param * sparse.eye(self.num_factors)solve_vecs = np.zeros((num_solve, self.num_factors))t = time.time()for i in xrange(num_solve):if user:counts_i = self.counts[i].toarray()else:#假设要求item_vec,counts_i为counts中的第i列的转置counts_i = self.counts[:, i].T.toarray()''' 原论文中c_ui=1+alpha*r_ui,可是在计算Y’CuY时为了减少时间复杂度,利用了Y'CuY=Y'Y+Y'(Cu-I)Y,因为Cu是对角矩阵,其元素为c_ui，即1+alpha*r_ui。所以Cu-I也就是对角元素为alpha*r_ui的对角矩阵'''CuI = sparse.diags(counts_i, [0])pu = counts_i.copy()#np.where(pu != 0)返回pu中元素不为0的索引，然后将这些元素赋值为1,不知道这里为什么要赋值为1?pu[np.where(pu != 0)] = 1.0YTCuIY = fixed_vecs.T.dot(CuI).dot(fixed_vecs)YTCupu = fixed_vecs.T.dot(CuI + eye).dot(sparse.csr_matrix(pu).T)xu = spsolve(YTY + YTCuIY + lambda_eye, YTCupu)solve_vecs[i] = xuif i % 1000 == 0:print 'Solved %i vecs in %d seconds' % (i, time.time() - t)t = time.time()return solve_vecs