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 time

def load_matrix(filename, num_users, num_items):
    t0 = time.time()
    counts = np.zeros((num_users, num_items))
    total = 0.0
    num_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:
            continue
        if item >= num_items:
            continue
        if count != 0:
            counts[user, item] = count
            total += count
            num_zeros -= 1
        if 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alpha
    alpha = num_zeros / total
    print ‘alpha %.2f‘ % alpha
    counts *= 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 counts


class ImplicitMF():

    def __init__(self, counts, num_factors=40, num_iterations=30,
                 reg_param=0.8):
        self.counts = counts
        self.num_users = counts.shape[0]
        self.num_items = counts.shape[1]
        self.num_factors = num_factors
        self.num_iterations = num_iterations
        self.reg_param = reg_param

    def 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_items
        num_solve = self.num_users if user else self.num_items
        num_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)] = 1.0
            YTCuIY = 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] = xu
            if i % 1000 == 0:
                print ‘Solved %i vecs in %d seconds‘ % (i, time.time() - t)
                t = time.time()
        return solve_vecs