熟悉numpy和pandas

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本文参考资料：

写在前面

看了gitbook那本书的第二章和第三章，自己正在复习高等数学、线性代数、概率、统计的知识，所以想用numpy和pandas把第二章和第三章的逻辑实现，书中的代码有点复杂。自从有了这个想法后，gitbook那本书上的源码我几乎都没有看过。

说实话，我看教程很少看别人的源码，当了解问题的逻辑后，我第一时间想到的是自己来实现，写完了代码后再看别人的代码，如果实在写不出来，那么才看别人的源码，看完了之后我也会自己重新理解逻辑再重新敲一遍。

所以，如果你对用numpy和pandas把第二章和第三章的逻辑实现这件事有兴趣，可以在自己实现以后，来看看我写的，欢迎批评建议。

本文需要的基础如下：

基于用户的推荐

基本的事实：两个人的品味如电影品味越相近，那么两个人对相应内容的评价也越相近
逻辑：根据两者对同样内容的评价，分析两者的相似程度，给对方进行推荐

  • 怎么判断相似？
    • 曼哈顿距离
    • 欧氏距离
    • 皮尔逊系数
    • 余弦定理
  • 怎么推荐？
    • 找到与之最相似的用户，推荐这位用户评价最高的内容
    • 找到与之比较相似的用户，加权处理

距离

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# -*- coding: utf-8 -*-
__author__ = 'duohappy'

import numpy as np
import pandas as pd


users = {"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}
        }
        
    
def manhattan_pd(user1, user2):
    
    # nan在sum过程中以0替代
    distance = np.abs(user1 - user2).sum()
    
    return distance
    
def topN_pd(user, users_df):
    
    users_df = users_df.drop(user.name, axis=1)  # 注意！users.df.drop() Return new object with labels in requested axis removed.
    # print(users_df)
    
    distances = []
    
    for other_user in users_df:
        distance = manhattan_pd(user, users_df[other_user])   # users_df[other_user] 和 users_df.other_user表达含义不同！
        distances.append((other_user, distance))
        
    distances = sorted(distances, key=lambda x:x[-1])
    
    return distances
    
def recommend_pd(user, users_df):
    
    distances = topN_pd(user, users_df)
    
    nearest_user = distances[0][0]

    # users_df[nearest_user] 得到距离user最近的用户
    # user.isnull() 得到user没有试过的选择
    # users_df[nearest_user][user.isnull()] 得到user没有试过的选择在nearest_user中的情况
    # users_df[nearest_user][user.isnull()].dropna() 丢弃nearest_user也没有试过的选择，剩下的就可以推荐了
    recommendations = users_df[nearest_user][user.isnull()].dropna()
    recommendations = recommendations.sort_values(ascending=False)
    
    return recommendations
    
if __name__ == '__main__':
    
    # print(manhattan(users["Angelica"], users["Bill"]))
    # print(topN('Angelica', users))
    print(recommend('Bill', users))
    
    users_df = pd.DataFrame(users)
    # print(users_df)
    # manhattan_pd(users_df.Angelica, users_df.Bill)
    # print(topN_pd(users_df.Angelica, users_df))
    recommend_pd(users_df.Bill, users_df)

皮尔逊系数

为什么出现皮尔逊系数？那本书写的很清楚，建议一读

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# -*- coding: utf-8 -*-
__author__ = 'duohappy'


import numpy as np
import pandas as pd


users = {"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}
        }

def pearson(user1, user2):
    
    # 贸然地给填充0是不可行的，别人就没有看过这部电影，怎么轻易断定别人的评分是0
    # user1 = user1.fillna(0)
    # user2 = user2.fillna(0)
    
    # user1.isnull()得到包含True/False的Series
    # user1[user1.isnull()]得到值为NaN的Series
    # user1[user1.isnull()].index得到Series的index
    # list
    # 为了作并集操作，set
    user1_nan_labels = set(list(user1[user1.isnull()].index))
    user2_nan_labels = set(list(user2[user2.isnull()].index))
    all_nan_labels = user1_nan_labels | user2_nan_labels
    
    # 只用两者都有的内容计算
    # 丢弃所有的NaN
    user1 = user1.drop(all_nan_labels)
    user2 = user2.drop(all_nan_labels)
    
    user1_sub_mean = user1 - user1.mean()
    user2_sub_mean = user2 - user2.mean()
    
    # 向量的内积
    num = (user1_sub_mean).dot(user2_sub_mean)
    # 向量内积
    den = np.sqrt(user1_sub_mean.dot(user1_sub_mean)) * np.sqrt(user2_sub_mean.dot(user2_sub_mean))
    
    return num/den
    
        
if __name__ == '__main__':
    
    users_df = pd.DataFrame(users)
    
    pearson(users_df.Angelica, users_df.Bill)

    # print(users_df.Angelica)
    # print(users_df.Hailey)

余弦定理

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# -*- coding: utf-8 -*-
__author__ = 'duohappy'


import numpy as np
import pandas as pd

users = {"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}
        }
        
        
def cosine(user1, user2):
    # 余弦定理用0代替空值
    user1 = user1.fillna(0)
    user2 = user2.fillna(0)
    
    num = user1.dot(user2)
    den = np.sqrt(user1.dot(user1))*np.sqrt(user2.dot(user2))
    
    return num/den
    
if __name__ == '__main__':
    users_df = pd.DataFrame(users)
    
    cosine(users_df.Angelica, users_df.Veronica)

K最邻近算法

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# -*- coding: utf-8 -*-
__author__ = 'duohappy'


import numpy as np
import pandas as pd

from pearson_coefficient import pearson
        
users={'Lisa Rose': {'Lady in the Water': 2.5, 'Snakes on a Plane': 3.5,
 'Just My Luck': 3.0, 'Superman Returns': 3.5, 'You, Me and Dupree': 2.5, 
 'The Night Listener': 3.0},
'Gene Seymour': {'Lady in the Water': 3.0, 'Snakes on a Plane': 3.5, 
 'Just My Luck': 1.5, 'Superman Returns': 5.0, 'The Night Listener': 3.0, 
 'You, Me and Dupree': 3.5}, 
'Michael Phillips': {'Lady in the Water': 2.5, 'Snakes on a Plane': 3.0,
 'Superman Returns': 3.5, 'The Night Listener': 4.0},
'Claudia Puig': {'Snakes on a Plane': 3.5, 'Just My Luck': 3.0,
 'The Night Listener': 4.5, 'Superman Returns': 4.0, 
 'You, Me and Dupree': 2.5},
'Mick LaSalle': {'Lady in the Water': 3.0, 'Snakes on a Plane': 4.0, 
 'Just My Luck': 2.0, 'Superman Returns': 3.0, 'The Night Listener': 3.0,
 'You, Me and Dupree': 2.0}, 
'Jack Matthews': {'Lady in the Water': 3.0, 'Snakes on a Plane': 4.0,
 'The Night Listener': 3.0, 'Superman Returns': 5.0, 'You, Me and Dupree': 3.5},
'Toby': {'Snakes on a Plane':4.5,'You, Me and Dupree':1.0,'Superman Returns':4.0}}
        
def topN(user, users_df):
    
    # 和其他用户比较
    # 丢弃掉user.name这一列
    users_df = users_df.drop(user.name, axis=1)
    # print(users_df) # 函数内重新定义了一个变量users_df，而在外部的users_df并没有改变，users_df.drop()方法并不会改变本身的users_df
    
    similarity = []  # 相似度
    
    for other_user in users_df:
        positive_sim = pearson(users_df[other_user], user)  # 其他人和user的皮尔逊系数
        if positive_sim > 0: # 只保留正相关系数
            similarity.append((other_user, positive_sim))
        
    similarity = sorted(similarity, key=lambda x: x[-1], reverse= True)  # 排序
    
    print(similarity)
    
    return similarity
    
def recommend(user, users_df):
    sim = topN(user, users_df)  # 得到相似度
    
    # sim = [sim[0]]
    all_choice = []  # 所有的选择
    
    for item in sim:
        all_choice.extend(list(users_df[item[0]].index))  # 得到其他人的所有选择
    
    available_choice = set(all_choice) - set(list(user.dropna().index))  # 得到自己没有做过的选择，注意一定要dropna
    
    print(available_choice)
    
    recom = []  # 推荐
    
    # 每一个没有做过的选择
    for choice in available_choice: 
        sum_score = 0  # 总的评分
        sum_sim = 0  # 总的相似度
        for item in sim: 　# 每一个其他用户
            score = users_df.loc[choice, item[0]] # 其他用户的选择 评分
            if not np.isnan(score):  # 如果这个评分不是nan的话
                sum_score += item[-1] * users_df.loc[choice, item[0]]  # 计算总分
                sum_sim += item[-1]  # 总的相似度

        recom.append((choice, sum_score/sum_sim))  # sum_score/sum_sim
        
    recom = sorted(recom, key=lambda x: x[-1], reverse=True)
    
    
    return recom
            
    
    
if __name__ == '__main__':
    users_df = pd.DataFrame(users)
    
    # topN(users_df.Sam, users_df)
    
    recommend(users_df.Toby, users_df)

基于物品的推荐

基本事实：两个物品很相近，那么同一用户对这两个物品的评价很可能也相近

修正的余弦定理

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# -*- coding: utf-8 -*-
__author__ = 'duohappy'


import numpy as np
import pandas as pd

users = {"David": {"Imagine Dragons": 3, "Daft Punk": 5,
                    "Lorde": 4, "Fall Out Boy": 1},
          "Matt": {"Imagine Dragons": 3, "Daft Punk": 4,
                   "Lorde": 4, "Fall Out Boy": 1},
          "Ben": {"Kacey Musgraves": 4, "Imagine Dragons": 3,
                  "Lorde": 3, "Fall Out Boy": 1},
          "Chris": {"Kacey Musgraves": 4, "Imagine Dragons": 4,
                    "Daft Punk": 4, "Lorde": 3, "Fall Out Boy": 1},
          "Tori": {"Kacey Musgraves": 5, "Imagine Dragons": 4,
                   "Daft Punk": 5, "Fall Out Boy": 3}}
                   
               
def cosine(item1, item2, users_df):
    # item2.notnull() item2不为nan的位置
    # item1[item2.notnull()] 取出item1中对应的value
    # item1[item2.notnull()].dropna() 丢弃item1中特有的nan
    # item1 = item1[item2.notnull()].dropna()，这个种写法会影响下面一条语句
    # item1_dropna = item1[item2.notnull()].dropna()
    # item2_dropna = item2[item1.notnull()].dropna()
    
    # 用&而不是and
    # 如果item1和item2中对应位置有一个nan，那么得到的bool array对应位置就是False
    all_not_nan = item1.notnull().values & item2.notnull().values
    
    item1 = item1[all_not_nan]
    item2 = item2[all_not_nan]
    
    
    # item1.index和item2.index包含同样的内容
    # 必须要dropna后计算mean
    data = [users_df[name].dropna().mean() for name in item1.index]
    # 组成一个对应的Series
    item1_mean = pd.Series(data, item1.index)
    item2_mean = pd.Series(data, item2.index)
    
    item1_sub_mean = item1 - item1_mean
    item2_sub_mean = item2 - item2_mean
    
    # print(item1_sub_mean)
    # print(item2_sub_mean)
    
    num = item1_sub_mean.dot(item2_sub_mean)
    den = np.sqrt(item1_sub_mean.dot(item1_sub_mean)) * np.sqrt(item2_sub_mean.dot(item2_sub_mean))
    
    # print(num / den)
    
    return num/den
    

# 得到修正的余弦相似度矩阵
def cosineMatrix(users_df):
    
    # 构造这个矩阵的大致样子
    cosine_matrix = pd.DataFrame(np.zeros([5, 5]), index=users_df.index, columns=users_df.index)
    
    # 定义一个索引i
    i = 0
    
    # 遍历行列，计算元素具体值
    for row in users_df.index:
        i += 1
        # 为了避免重复计算，使用切片
        for column in users_df.index[:i]:
            if row == column:
                cosine_matrix.loc[row, column] = 1
                continue
            cosine_matrix.loc[row, column] = cosine(users_df.loc[row], users_df.loc[column], users_df)
            cosine_matrix.loc[column, row] = cosine_matrix.loc[row, column]
            
    cosine_matrix.to_csv('./tmp.csv')
    
    return cosine_matrix
    
    
def predict_score(user, item, users_df):
    MAX = 5
    MIN = 1
    
    all_nan = user.notnull()
    
    user = user[all_nan]
    cosine_item = cosineMatrix(users_df)[item.name][all_nan]
    
    # print(user)
    # print(cosine_item)
    
    user = (2*(user - MIN) - (MAX - MIN))/(MAX - MIN)
    # print(user)
    
    num = user.dot(cosine_item)
    den = np.abs(cosine_item).sum()
    
    nr = num/den
    r = 1/2*((nr + 1) * (MAX - MIN)) + MIN
    
    return nr, r
    
    
if __name__ == '__main__':
    users_df = pd.DataFrame(users)
    
    print(users_df)
    # print(users_df.loc['Daft Punk'], users_df.loc['Fall Out Boy'])
    
    # DataFrame取一行使用loc
    # cosine(users_df.loc['Kacey Musgraves'], users_df.loc['Imagine Dragons'], users_df)
    
    # cosineMatrix(users_df)
    predict_score(users_df.David, users_df.loc['Kacey Musgraves'], users_df)

slope one

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# -*- coding: utf-8 -*-
__author__ = 'duohappy'


import numpy as np
import pandas as pd

users = {"Amy": {"Taylor Swift": 4, "PSY": 3, "Whitney Houston": 4},
          "Ben": {"Taylor Swift": 5, "PSY": 2},
          "Clara": {"PSY": 3.5, "Whitney Houston": 4},
          "Daisy": {"Taylor Swift": 5, "Whitney Houston": 3}}
          
def slope_one_func(item1, item2):

    all_not_nan = item1.notnull() & item2.notnull()
    
    item1 = item1[all_not_nan]
    item2 = item2[all_not_nan]
    
    card_s = len(item1)
    
    dev = ((item1 - item2) / card_s).sum()
    
    return dev
    
def slop_one_matrix(users_df):
    
    matrix = pd.DataFrame(np.zeros((3, 3)), index=users_df.index, columns=users_df.index)
    
    i = 0
    for row in users_df.index:
        i += 1
        for column in users_df.index[:i]:
            if row == column:
                matrix.loc[row, column] = 0
                continue
            matrix.loc[row, column] = slope_one_func(users_df.loc[row], users_df.loc[column])
            matrix.loc[column, row] = -matrix.loc[row, column]
            
    
    return matrix
    
    
    
if __name__ == '__main__':
    
    users_df = pd.DataFrame(users)
    
    print(users_df)
    # slope_one_func(users_df.loc['PSY'], users_df.loc['Taylor Swift'])
    
    slop_one_matrix(users_df)

写在后面

本文仅仅为了熟悉numpy和pandas用法，至于推荐系统嘛，我还没有学太深，先不写。太基础的内容，我不会写，因为网上太多资料了。我会坚持写点符合自己个性的文章。

越来越发现业务知识的重要性，不然分析来分析去，分析出一堆数，也不知道分析的意义。定性分析先行，定量分析后行。

您的支持将鼓励我继续创作！