【Spark机器学习速成宝典】推荐引擎——协同过滤

目录

  推荐模型的分类

  ALS交替最小二乘算法:显式矩阵分解

  Spark Python代码:显式矩阵分解

  ALS交替最小二乘算法:隐式矩阵分解

  Spark Python代码:隐式矩阵分解


推荐模型的分类

  最为流行的两种方法是基于内容的过滤协同过滤

  基于内容的过滤

  比如用户A买了商品A,商品B与商品A相似(这个相似是基于商品内部的属性,比如“非常好的协同过滤入门文章”和“通俗易懂的协同过滤入门教程”比较相似),那么就能将商品B推荐给用户。

  协同过滤

  利用的是训练数据是大量用户对商品的评分,即(userID,productID,score)。称得上最经典最常用的推荐算法。协同过滤又可细分为基于用户的推荐基于物品的推荐

  基于用户的推荐

  简单解释就是“志趣相投”

  以商品为维度,寻找相似用户。就能给用户A推送他的相似用户买过的商品。

  基于物品的推荐

  简单解释就是“物以类聚”

  以用户为维度,寻找相似商品。比如用户A买了个商品A,那就能推荐与商品A相似的商品。

 返回目录

ALS交替最小二乘算法:显式矩阵分解 

  ALS(Alternating Least squares)算法是用来求解协同过滤模型的重要算法。在训练集中我们有用户对商品打分的矩阵:

  问题:用户u5给商品v4的打分大概是多少呢?这就是协同过滤算法要做的事情,即求出每一个格子里的值,这个值就是某用户对某商品的评分。那如何做呢?答:矩阵分解。

  首先,将用户评分矩阵Am×n拆成用户特征矩阵Um×k乘以商品特征矩阵Vk×n的形式:

  

  其中A=U×V,如果得到了UV,那么用户评分矩阵中的A(i,j),可由向量U(i,:)点乘向量V(:,j)计算得到。

  用两个小矩阵 Um×k 和 Vn×k 的乘积近似等价Am×n。这样,整个系统的自由度从o(mn)降到o((m+n)k)。代价函数可以设置为:矩阵A中每个元素与重构矩阵U×V之间的每个元素的误差平方和,即:

  为防止过拟合,一般加入L2正则化项:

  但是这个损失函数不是凸的,而且变量互相耦合在一起,不宜求解。那怎么办? 答:使用ALS算法。

  ALS算法的思想是:将 UV 固定其一,这个最优化问题立刻变成一个凸的可拆分的问题,求解过程就是基于最小二乘法的最优化问题。

  ALS算法完整求解过程:先随机生成 U(0),然后固定 U(0)求解 V(0),再固定 V(0)求解 U(1)这样交替进行下去,由于总体问题的非凸性,ALS并不保证收敛到全局最优解,但在实际应用中迭代10次,便能训练出较好的模型。

  最终我们可以得到这三个矩阵:

  通过这些矩阵便可以做以下事情:

  1.为用户推荐潜在感兴趣的TopK个商品

  2.为商品寻找潜在的TopK个用户

  3.寻找相似用户

  4.寻找相似商品

  模型训练中涉及到的参数调节的准则:

  rank:对应ALS模型中的因子个数,也就是在低阶近似矩阵中的隐含特征个数。因子个数一般越多越好。但它也会直接影响模型训练和保存时所需的内存开销,尤其是在用户和物品很多的时候。因此实践中该参数常作为训练效果与系统开销之间的调节参数。通常,其合理取值为10到200。

  iterations:对应运行时的迭代次数。ALS能确保每次迭代都能降低评级矩阵的重建误差,但一般经少数次迭代后ALS模型便已能收敛为一个比较合理的好模型。这样,大部分情况下都没必要迭代太多次(10次左右一般就挺好)。

  lambda:该参数控制模型的正则化过程,从而控制模型的过拟合情况。其值越高,正则化越严厉。该参数的赋值与实际数据的大小、特征和稀疏程度有关。和其他的机器学习模型一样,正则参数应该通过用非样本的测试数据进行交叉验证来调整。

 返回目录

Spark Python代码:显式矩阵分解 

# -*-coding=utf-8 -*-  
from pyspark import SparkConf, SparkContext
from pyspark.mllib.recommendation import ALS, MatrixFactorizationModel, Rating
sc = SparkContext("local")


# 加载和解析数据
data = sc.textFile("test.data")
ratings = data.map(lambda l: l.split(',')).map(lambda l: Rating(int(l[0]), int(l[1]), float(l[2])))

# 使用交替最小二乘算法训练推荐模型
rank = 3
numIterations = 10
model = ALS.train(ratings, rank, numIterations)

# 打印用户特征矩阵
print "用户特征矩阵:"
a = model.userFeatures().collect()
for i in a:
    print str(i[1][0])[:5] + "," + str(i[1][1])[:5] + "," + str(i[1][2])[:5]

# 打印物品特征矩阵
print "物品特征矩阵:"
b = model.productFeatures().collect()
for i in b:
    print str(i[1][0])[:5] + "," + str(i[1][1])[:5] + "," + str(i[1][2])[:5]

# 打印用户评分矩阵
print "用户评分矩阵:"
res = model.recommendProductsForUsers(9) #为每一个用户推荐num个商品
def a(userdata):
    lst = []
    userid = int(userdata[0])
    lst.append(userid)
    for i in userdata[1]:
        product = str(i.product) + ":" + str(i.rating)[:5]
        lst.append(product)
    return lst
res = res.map(a).collect()
res.sort(key=lambda x:x[0]) 
for line in res:
    line = line[1:]
#     print line
    j = [i.split(":") for i in line]
    j.sort(key=lambda x:x[0]) 
    l=""
    for k in j:
        l += k[1] + ","
    print l

a = model.predict(user=1, product=2)
print "预测user1与product2的兴趣度:"
print a

a = model.recommendUsers(product=2, num=2)
print "为商品2寻找潜在的TopK个用户:"
print a

a = model.recommendProducts(user=1, num=3)
print "为user1推荐潜在感兴趣的TopK个商品:"
print a

a = model.recommendProductsForUsers(num=3).collect()
print "为所有用户推荐潜在感兴趣的TopK个商品:" 
for i in a:
    print i
    
a = model.recommendUsersForProducts(num=3).collect()
print "为所有商品寻找潜在的TopK个用户:" 
for i in a:
    print i

'''res
17/12/23 14:50:00 WARN BLAS: Failed to load implementation from: com.github.fommil.netlib.NativeSystemBLAS
17/12/23 14:50:00 WARN BLAS: Failed to load implementation from: com.github.fommil.netlib.NativeRefBLAS
17/12/23 14:50:01 WARN LAPACK: Failed to load implementation from: com.github.fommil.netlib.NativeSystemLAPACK
17/12/23 14:50:01 WARN LAPACK: Failed to load implementation from: com.github.fommil.netlib.NativeRefLAPACK
17/12/23 14:50:01 WARN Executor: 1 block locks were not released by TID = 29:
[rdd_210_0]
17/12/23 14:50:01 WARN Executor: 1 block locks were not released by TID = 30:
[rdd_211_0]
17/12/23 14:50:01 WARN Executor: 1 block locks were not released by TID = 31:
[rdd_210_0]
17/12/23 14:50:01 WARN Executor: 1 block locks were not released by TID = 32:
[rdd_211_0]
用户特征矩阵:
0.575,-0.52,-0.76
0.165,1.110,-0.51
0.597,0.941,-0.07
0.516,0.430,-0.52
-0.28,0.113,1.037
0.385,-0.34,-1.07
0.236,0.805,-0.66
物品特征矩阵:
2.467,1.397,-7.93
1.896,3.044,-0.17
3.575,3.367,-3.25
1.584,2.145,-0.48
-1.53,3.854,5.902
1.358,-0.08,-4.12
0.121,6.417,0.299
3.823,-3.44,-5.18
2.387,3.749,-0.29
用户评分矩阵:
6.801,-0.36,2.801,0.163,-7.44,4.000,-3.52,7.996,-0.36,
6.028,3.787,6.000,2.895,1.003,2.241,6.992,-0.53,4.712,
3.357,4.013,5.539,3.002,2.295,1.025,6.095,-0.58,4.979,
6.020,2.383,4.995,1.996,-2.21,2.815,2.670,3.197,3.002,
-8.78,-0.38,-4.02,-0.71,6.996,-4.67,1.001,-6.86,-0.56,
9.004,-0.13,3.707,0.386,-8.28,4.986,-2.51,8.252,-0.06,
6.998,3.020,5.727,2.427,-1.19,2.997,4.996,1.587,3.783,
预测user与product的兴趣度:
-0.364804845726
为商品2寻找潜在的TopK个用户:
[Rating(user=3, product=2, rating=4.013395373544592), Rating(user=2, product=2, rating=3.7876436839883194)]
为user1推荐潜在感兴趣的TopK个商品:
[Rating(user=1, product=8, rating=7.996952478409568), Rating(user=1, product=1, rating=6.801415502963593), Rating(user=1, product=6, rating=4.000096674124945)]
为所有用户推荐潜在感兴趣的TopK个商品:
(4, (Rating(user=4, product=1, rating=6.020880222950041), Rating(user=4, product=3, rating=4.995487526087871), Rating(user=4, product=8, rating=3.1971086869642846)))
(1, (Rating(user=1, product=8, rating=7.996952478409568), Rating(user=1, product=1, rating=6.801415502963593), Rating(user=1, product=6, rating=4.000096674124945)))
(6, (Rating(user=6, product=1, rating=9.004180651398052), Rating(user=6, product=8, rating=8.25221831618579), Rating(user=6, product=6, rating=4.986725295866833)))
(3, (Rating(user=3, product=7, rating=6.095926975521101), Rating(user=3, product=3, rating=5.5399318710354155), Rating(user=3, product=9, rating=4.979915488571464)))
(7, (Rating(user=7, product=1, rating=6.998396543388047), Rating(user=7, product=3, rating=5.72731910674824), Rating(user=7, product=7, rating=4.9965585464819355)))
(5, (Rating(user=5, product=5, rating=6.996394158606613), Rating(user=5, product=7, rating=1.0018132801060546), Rating(user=5, product=2, rating=-0.386246390781805)))
(2, (Rating(user=2, product=7, rating=6.9928960101812), Rating(user=2, product=1, rating=6.02809652321465), Rating(user=2, product=3, rating=6.000155003043972)))
为所有商品寻找潜在的TopK个用户:
(4, (Rating(user=3, product=4, rating=3.0029246134929766), Rating(user=2, product=4, rating=2.8953672065005556), Rating(user=7, product=4, rating=2.427766056801616)))
(1, (Rating(user=6, product=1, rating=9.004180651398052), Rating(user=7, product=1, rating=6.998396543388047), Rating(user=1, product=1, rating=6.801415502963593)))
(6, (Rating(user=6, product=6, rating=4.986725295866833), Rating(user=1, product=6, rating=4.000096674124945), Rating(user=7, product=6, rating=2.997989324292092)))
(3, (Rating(user=2, product=3, rating=6.000155003043972), Rating(user=7, product=3, rating=5.72731910674824), Rating(user=3, product=3, rating=5.5399318710354155)))
(7, (Rating(user=2, product=7, rating=6.9928960101812), Rating(user=3, product=7, rating=6.095926975521101), Rating(user=7, product=7, rating=4.9965585464819355)))
(9, (Rating(user=3, product=9, rating=4.979915488571464), Rating(user=2, product=9, rating=4.712732268716017), Rating(user=7, product=9, rating=3.7833941032831433)))
(8, (Rating(user=6, product=8, rating=8.25221831618579), Rating(user=1, product=8, rating=7.996952478409568), Rating(user=4, product=8, rating=3.1971086869642846)))
(5, (Rating(user=5, product=5, rating=6.996394158606613), Rating(user=3, product=5, rating=2.295168542063017), Rating(user=2, product=5, rating=1.0030496463981144)))
(2, (Rating(user=3, product=2, rating=4.013395373544592), Rating(user=2, product=2, rating=3.7876436839883194), Rating(user=7, product=2, rating=3.0206127312499014)))
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'''

  

  test.data数据:
1,6,4.0
1,8,8.0
2,3,6.0
2,5,1.0
2,7,7.0
3,2,4.0
3,4,3.0
3,9,5.0
4,3,5.0
4,4,2.0
4,9,3.0
5,5,7.0
5,7,1.0
6,1,9.0
6,6,5.0
7,1,7.0
7,6,3.0
7,7,5.0

 返回目录

ALS交替最小二乘算法:隐式矩阵分解 

  在显式矩阵分解的训练集中,我们有用户对商品的评级,用户评分的高低决定了用户对商品的感兴趣的程度,假设有10个等级,那么评1级就是意味着讨厌,评10级意味着非常喜欢。

  但是,有些时候我们没有用户显式的打分数据,而是有用户的行为数据。

  比如:用户对某一个电影观看多次,可以说这个用户喜欢这个电影,而且置信度很高;如果用户对某个电影观看了一次,我们也可以说用户喜欢这个电影,只是置信度低。

  下面介绍:隐式矩阵分解。

  首先我们有数据集:

  先将数据集依照公式:

  ,      

  拆成2个矩阵:二元偏好矩阵P、信心权重矩阵C(这里取α=0.1,α也是一个正则化参数)

 

   代价函数是:

  求解的方法和显式矩阵分解一样。  

 返回目录

Spark Python代码:隐式矩阵分解 

# -*-coding=utf-8 -*-  
from pyspark import SparkConf, SparkContext
from pyspark.mllib.recommendation import ALS, MatrixFactorizationModel, Rating
sc = SparkContext("local")


# 加载和解析数据
data = sc.textFile("test.data")
ratings = data.map(lambda l: l.split(',')).map(lambda l: Rating(int(l[0]), int(l[1]), float(l[2])))

# 使用交替最小二乘算法训练推荐模型
rank = 3
numIterations = 10
model = ALS.train(ratings, rank, numIterations)
# 打印用户评分矩阵
def a(userdata):
    lst = []
    userid = int(userdata[0])
    lst.append(userid)
    for i in userdata[1]:
        product = str(i.product) + ":" + str(i.rating)[:5]
        lst.append(product)
    return lst
    
model = ALS.trainImplicit(ratings, rank, numIterations, alpha=0.1)

res = model.recommendProductsForUsers(9) #为每一个用户推荐num个商品
res = res.map(a).collect()
res.sort(key=lambda x:x[0]) 
for line in res:
    line = line[1:]
#     print line
    j = [i.split(":") for i in line]
    j.sort(key=lambda x:x[0]) 
    l=""
    for k in j:
        l += k[1] + ","
    print l

'''
0.674,0.072,-0.15,0.054,-0.23,0.891,0.089,0.432,0.054,
0.091,-0.03,0.771,0.125,0.999,-0.04,1.085,-0.18,0.126,
-0.01,0.601,0.501,0.989,-0.17,0.055,-0.19,0.089,0.991,
-0.04,0.547,0.725,0.962,0.171,-0.04,0.144,-0.01,0.964,
0.128,-0.14,0.493,-0.10,0.814,0.019,0.913,-0.13,-0.10,
0.778,0.031,-0.09,0.005,-0.09,0.997,0.284,0.454,0.005,
0.887,-0.05,0.158,-0.05,0.330,1.058,0.787,0.404,-0.05,
'''

 返回目录

原文地址:https://www.cnblogs.com/itmorn/p/8074779.html