'''基于物品的协同推荐矩阵数据说明：1.修正的余弦相似度是一种基于模型的协同过滤算法。我们前面提过，这种算法的优势之一是扩展性好，对于大数据量而言，运算速度快、占用内存少。2.用户的评价标准是不同的，比如喜欢一个歌手时有些人会打4分，有些打5分；不喜欢时有人会打3分，有些则会只给1分。修正的余弦相似度计算时会将用户对物品的评分减去用户所有评分的均值，从而解决这个问题。如何预测用户对给定物品的打分？一、基于用户协同方法1：用户之间的距离/相似度（K近邻算法）二、基于物品协同方法1：物品之间的相似度（修正的余弦相似度）方法2：物品之间的差异值（加权Slope One算法）由于使用的是用户产生的各种数据，因此又称为社会化过滤算法。协同过滤会遇到的种种问题，包括数据的稀疏性和算法的可扩展性。此外，协同过滤算法倾向于推荐那些已经很流行的物品。这类推荐系统会让流行的物品更为流行，冷门的物品更无人问津。-- Daniel Fleder & Kartik Hosanagar 2009 《 推荐系统对商品分类的影响》'''import pandas as pdfrom io import StringIO#数据类型一：条目（用户、商品、打分）（避免巨型稀疏矩阵）csv_txt = '''"David","Imagine Dragons",3"David","Daft Punk",5"David","Lorde",4"David","Fall Out Boy",1"Matt","Imagine Dragons",3"Matt","Daft Punk",4"Matt","Lorde",4"Matt","Fall Out Boy",1"Ben","Kacey Musgraves",4"Ben","Imagine Dragons",3"Ben","Lorde",3"Ben","Fall Out Boy",1"Chris","Kacey Musgraves",4"Chris","Imagine Dragons",4"Chris","Daft Punk",4"Chris","Lorde",3"Chris","Fall Out Boy",1"Tori","Kacey Musgraves",5"Tori","Imagine Dragons",4"Tori","Daft Punk",5"Tori","Fall Out Boy",3'''#数据类型一：条目（用户、商品、打分）（避免巨型稀疏矩阵）#根据《data minning guide》第85页的users2数据csv_txt2 = '''"Amy","Taylor Swift",4"Amy","PSY",3"Amy","Whitney Houston",4"Ben","Taylor Swift",5"Ben","PSY",2"Clara","PSY",3.5"Clara","Whitney Houston",4"Daisy","Taylor Swift",5"Daisy","Whitney Houston",3'''df = None#方式一：加载csv数据def load_csv_txt():    global df    #df = pd.read_csv(StringIO(csv_txt2), header=None, names=['user','goods','rate'])    df = pd.read_csv("BX-Book-Ratings.csv", header=None, names=['user','goods','rate'], sep=';')            print('测试：读取数据')load_csv_txt()#load_json_txt()#=======================================# 注意：不需要build_xy#=======================================# 计算两个物品相似度def computeSimilarity(goods1, goods2):    '''根据《data minning guide》第71页的公式s(i,j)'''    df1 = df.ix[df['goods'].isin([goods1]), ['user','rate']]    df2 = df.ix[df['goods'].isin([goods2]), ['user','rate']]        df3 = pd.merge(df1, df2, on='user', how='inner') #只保留两种商品都被同一用户评过分的商品的评分        # 注意：先构造了一个只有index=df3['user']的pd.Series，目的是方便下一步的减    mean = pd.Series(df3['user'].tolist(), index=df3['user']).apply(lambda x:df.ix[df['user']==x, 'rate'].mean())        # 每行的用户评分都减去了该用户的平均评分    df3.index = df3['user']    df3['rate_x'] = df3['rate_x'] - mean    df3['rate_y'] = df3['rate_y'] - mean        # 返回修正的余弦相似度    return sum(df3['rate_x'] * df3['rate_y']) / (sum(df3['rate_x']**2) * sum(df3['rate_y']**2))**0.5 # merge之后默认的列名：rate_x，rate_y        # csv_txt#print('\n测试：计算Kacey Musgraves与Imagine Dragons的相似度')#print(computeSimilarity("Kacey Musgraves","Imagine Dragons"))#为了让公式的计算效果更佳，对物品的评价分值最好介于-1和1之间def rate2newrate(rate):    '''根据《data minning guide》第76页的公式NR(u,N)'''    ma, mi = df['rate'].max(), df['rate'].min()    return (2*(rate - mi) - (ma - mi))/(ma - mi)        #已知rate2newrate求newrate2ratedef newrate2rate(new_rate):    '''根据《data minning guide》第76页的公式R(u,N)'''    ma, mi = df['rate'].max(), df['rate'].min()    return (0.5 * (new_rate + 1) * (ma - mi)) + mi    #print('\n测试：计算3的new_rate值')#print(rate2newrate(3))#print('\n测试：计算0.5的rate值')#print(newrate2rate(0.5))# 预测给定用户对物品的可能评分(对评分进行了修正/还原)def p2(user, goods):    '''根据《data minning guide》第75页的公式p(u,i)'''    #assert pd.isnull(df.ix[df['user']==user & df['goods']==goods, 'rate']) # 必须用户对给定物品尚未评分        #用户对其它所有已打分物品的打分数据    df2 = df.ix[df['user']==user, ['goods','rate']]    s1 = pd.Series(df2['rate'].tolist(), index=df2['goods'])    s1 = s1.apply(lambda x:rate2newrate(x)) #修正    s2 = s1.index.to_series().apply(lambda x:computeSimilarity(x, goods)) #打分物品分别与给定物品的相似度        return newrate2rate(sum(s1 * s2) / sum(abs(s2)))#还原# csv_txt#print('\n测试：预测David对Kacey Musgraves的打分（修正）')#print(p2("David","Kacey Musgraves"))#==================================# 下面是Slope One算法## 两个步骤：# 1. 计算差值# 2. 预测用户对尚未评分物品的评分#==================================# 1.计算两物品之间的差异def dev(goods1, goods2):    '''根据《data minning guide》第80页的公式dev(i,j)'''    #s = (df[goods1] - df[goods2]).dropna()        df1 = df.ix[df['goods'].isin([goods1]), ['user','rate']]    df2 = df.ix[df['goods'].isin([goods2]), ['user','rate']]    print('df1\n', df1)    print('df2\n', df2)        df3 = pd.merge(df1, df2, on='user', how='inner') #只保留两个物品都有评分的用户的评分    print(df3)    print(df3.shape[0])        d = sum(df3['rate_x'] - df3['rate_y'])/df3.shape[0] # 差异值    return d, df3.shape[0] #返回差异值，及权值（同时对两个物品打分的人数）# csv_txt#print('\n测试：计算Kacey Musgraves与Imagine Dragons的分数差异')#print(dev("Kacey Musgraves","Imagine Dragons"))#计算所有两两物品之间的评分差异，得到方阵pd.DataFrame（行对列）def get_dev_table():    '''根据《data minning guide》第87页的表'''    #goods_names = df.columns.tolist()    goods_names = df['goods'].unique().tolist()    df2 = pd.DataFrame(.0, index=goods_names, columns=goods_names) #零方阵        for i,goods1 in enumerate(goods_names):        for goods2 in goods_names[i+1:]:            d, _ = dev(goods1, goods2) # 注意：只取了物品差异值            df2.ix[goods1, goods2] = d            df2.ix[goods2, goods1] = -d # 对称的位置取反        return df2        #print('\n测试：计算所有两两物品之间的评分差异表')#print(get_dev_table())#预测某用户对给定物品的评分# 加权Slope One算法def slopeone(user, goods):    '''根据《data minning guide》第82页的公式p(u,j)'''    df1 = df.ix[df['user'].isin([user]), ['goods', 'rate']]        s1 = pd.Series(df1['rate'].tolist(), index=df1['goods']) # 用户对已打分物品的打分数据        s2 = s1.index.to_series().apply(lambda x:dev(goods, x)) # 待打分物品与已打分物品的差异值及权值        s3 = s2.apply(lambda x:x[0]) #差异值    s4 = s2.apply(lambda x:x[1]) #权值    #print(s1, s3, s4)        return sum((s1 + s3) * s4)/sum(s4)        #print('\n测试：加权Slope One算法，预测用户Ben对物品Whitney Houston的评分')#print(slopeone('Ben', 'Whitney Houston')) # 3.375print('\n测试：加权Slope One算法，预测用户276744对书本0600570967的评分') #大数据量print(slopeone(276744, '0600570967'))