本文共 6213 字,大约阅读时间需要 20 分钟。
原书作者使用字典dict实现推荐算法,并且惊叹于18行代码实现了向量的余弦夹角公式。
我用pandas实现相同的公式只要3行。
特别说明:本篇笔记是针对矩阵数据,下篇笔记是针对条目数据。
'''基于用户的协同推荐矩阵数据'''import pandas as pdfrom io import StringIOimport json#数据类型一:csv矩阵(用户-商品)(适用于小数据量)csv_txt = '''"user","Blues Traveler","Broken Bells","Deadmau5","Norah Jones","Phoenix","Slightly Stoopid","The Strokes","Vampire Weekend""Angelica",3.5,2.0,,4.5,5.0,1.5,2.5,2.0"Bill",2.0,3.5,4.0,,2.0,3.5,,3.0"Chan",5.0,1.0,1.0,3.0,5,1.0,,"Dan",3.0,4.0,4.5,,3.0,4.5,4.0,2.0"Hailey",,4.0,1.0,4.0,,,4.0,1.0"Jordyn",,4.5,4.0,5.0,5.0,4.5,4.0,4.0"Sam",5.0,2.0,,3.0,5.0,4.0,5.0,"Veronica",3.0,,,5.0,4.0,2.5,3.0,'''#数据类型二:json数据(用户、商品、打分)json_txt = '''{"Angelica": {"Blues Traveler": 3.5, "Broken Bells": 2.0, "Norah Jones": 4.5, "Phoenix": 5.0, "Slightly Stoopid": 1.5, "The Strokes": 2.5, "Vampire Weekend": 2.0}, "Bill":{"Blues Traveler": 2.0, "Broken Bells": 3.5, "Deadmau5": 4.0, "Phoenix": 2.0, "Slightly Stoopid": 3.5, "Vampire Weekend": 3.0}, "Chan": {"Blues Traveler": 5.0, "Broken Bells": 1.0, "Deadmau5": 1.0, "Norah Jones": 3.0, "Phoenix": 5, "Slightly Stoopid": 1.0}, "Dan": {"Blues Traveler": 3.0, "Broken Bells": 4.0, "Deadmau5": 4.5, "Phoenix": 3.0, "Slightly Stoopid": 4.5, "The Strokes": 4.0, "Vampire Weekend": 2.0}, "Hailey": {"Broken Bells": 4.0, "Deadmau5": 1.0, "Norah Jones": 4.0, "The Strokes": 4.0, "Vampire Weekend": 1.0}, "Jordyn": {"Broken Bells": 4.5, "Deadmau5": 4.0, "Norah Jones": 5.0, "Phoenix": 5.0, "Slightly Stoopid": 4.5, "The Strokes": 4.0, "Vampire Weekend": 4.0}, "Sam": {"Blues Traveler": 5.0, "Broken Bells": 2.0, "Norah Jones": 3.0, "Phoenix": 5.0, "Slightly Stoopid": 4.0, "The Strokes": 5.0}, "Veronica": {"Blues Traveler": 3.0, "Norah Jones": 5.0, "Phoenix": 4.0, "Slightly Stoopid": 2.5, "The Strokes": 3.0}}'''df = None#方式一:加载csv数据def load_csv_txt(): global df df = pd.read_csv(StringIO(csv_txt), header=0, index_col="user")#方式二:加载json数据(把json读成矩阵)def load_json_txt(): global df df = pd.read_json(json_txt, orient='index') #测试:读取数据load_csv_txt()#load_json_txt()def build_xy(user_name1, user_name2): #df2 = df.ix[[user_name1, user_name2]].dropna(axis=1) #return df2.ix[user_name1], df2.ix[user_name2] bool_array = df.ix[user_name1].notnull() & df.ix[user_name2].notnull() return df.ix[user_name1, bool_array], df.ix[user_name2, bool_array]#曼哈顿距离def manhattan(user_name1, user_name2): x, y = build_xy(user_name1, user_name2) return sum(abs(x - y)) #欧几里德距离def euclidean(user_name1, user_name2): x, y = build_xy(user_name1, user_name2) return sum((x - y)**2)**0.5 #闵可夫斯基距离def minkowski(user_name1, user_name2, r): x, y = build_xy(user_name1, user_name2) return sum(abs(x - y)**r)**(1/r) #皮尔逊相关系数def pearson(user_name1, user_name2): x, y = build_xy(user_name1, user_name2) mean1, mean2 = x.mean(), y.mean() #分母 denominator = (sum((x-mean1)**2)*sum((y-mean2)**2))**0.5 return [sum((x-mean1)*(y-mean2))/denominator, 0][denominator == 0] #余弦相似度(数据的稀疏性问题,在文本挖掘中应用得较多)def cosine(user_name1, user_name2): x, y = build_xy(user_name1, user_name2) #分母 denominator = (sum(x*x)*sum(y*y))**0.5 return [sum(x*y)/denominator, 0][denominator == 0]metric_funcs = { 'manhattan': manhattan, 'euclidean': euclidean, 'minkowski': minkowski, 'pearson': pearson, 'cosine': cosine}#df.ix[["Angelica","Bill"]].dropna(axis=1)print(manhattan("Angelica","Bill"))#计算最近的邻居def computeNearestNeighbor(user_name, metric='pearson', k=3, r=2): ''' metric: 度量函数 k: 返回k个邻居 r: 闵可夫斯基距离专用 返回:pd.Series,其中index是邻居名称,values是距离 ''' if metric in ['manhattan', 'euclidean']: return df.drop(user_name).index.to_series().apply(metric_funcs[metric], args=(user_name,)).nsmallest(k) elif metric in ['minkowski']: return df.drop(user_name).index.to_series().apply(metric_funcs[metric], args=(user_name, r,)).nsmallest(k) elif metric in ['pearson', 'cosine']: return df.drop(user_name).index.to_series().apply(metric_funcs[metric], args=(user_name,)).nlargest(k) print(computeNearestNeighbor('Hailey', metric='pearson'))#向给定用户推荐(返回:pd.Series)def recommend(user_name): # 找到距离最近的用户名 nearest_username = computeNearestNeighbor(user_name).index[0] # 找出邻居评价过、但自己未曾评价的乐队(或商品) # 结果:index是商品名称,values是评分 return df.ix[nearest_username, df.ix[user_name].isnull() & df.ix[nearest_username].notnull()].sort_values()#为Hailey做推荐print(recommend('Hailey'))#向给定用户推荐def recommend2(user_name, metric='pearson', k=3, n=5, r=2): ''' metric: 度量函数 k: 根据k个最近邻居,协同推荐 r: 闵可夫斯基距离专用 n: 推荐的商品数目 返回:pd.Series,其中index是商品名称,values是加权评分 ''' # 找到距离最近的k个邻居 nearest_neighbors = computeNearestNeighbor(user_name, metric='pearson', k=k, r=r) # 计算权值 if metric in ['manhattan', 'euclidean', 'minkowski']: # 距离越小,越类似 nearest_neighbors = 1 / nearest_neighbors # 所以,取倒数(或者别的减函数,如:y=2**-x) elif metric in ['pearson', 'cosine']: # 距离越大,越类似 pass nearest_neighbors = nearest_neighbors / nearest_neighbors.sum() #已经变为权值(pd.Series) # 逐个邻居找出其评价过、但自己未曾评价的乐队(或商品)的评分,并乘以权值 neighbors_rate_with_weight = [] for neighbor_name in nearest_neighbors.index: # 每个结果:pd.Series,其中index是商品名称,values是评分(已乘权值) neighbors_rate_with_weight.append(df.ix[neighbor_name, df.ix[user_name].isnull() & df.ix[neighbor_name].notnull()] * nearest_neighbors[neighbor_name]) # 把邻居们的加权评分拼接成pd.DataFrame,按列累加,取最大的前n个商品的评分 return pd.concat(neighbors_rate_with_weight, axis=1).sum(axis=1, skipna=True).nlargest(n) #为Hailey做推荐print(recommend2('Hailey', metric='manhattan', k=3, n=5))#为Hailey做推荐print(recommend2('Hailey', metric='euclidean', k=3, n=5, r=2))#为Hailey做推荐print(recommend2('Hailey', metric='pearson', k=1, n=5))
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