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KMeans与深度学习自编码AutoEncoder结合提高聚类效

KMeans与深度学习自编码AutoEncoder结合提高聚类效

作者: 天善智能 | 来源:发表于2018-04-18 15:08 被阅读40次

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    这几天在做用户画像,特征是用户的消费商品的消费金额,原始数据(部分)是这样的:

    id goods_name goods_amount

    男士手袋 1882.0

    淑女装 2491.0

    女士手袋 345.0

    基础内衣 328.0

    商务正装 4985.0

    时尚 969.0

    女饰品 86.0

    专业运动 399.0

    童装(中大童) 2033.0

    男士配件 38.0

    我们看到同一个id下面有不同的消费记录,这个数据不能直接拿来用,写了python程序来进行处理:

    test.py#!/usr/bin/python

    #coding:utf-8

    #Author:Charlotte

    import pandas as pd

    import numpy as np

    import time

    #加载数据文件(你可以加载自己的文件,文件格式如上所示)

    x=pd.read_table('test.txt',sep = " ")

    #去除NULL值

    x.dropna()

    a1=list(x.iloc[:,0])

    a2=list(x.iloc[:,1])

    a3=list(x.iloc[:,2])

    #A是商品类别

    dicta=dict(zip(a2,zip(a1,a3)))

    A=list(dicta.keys())

    #B是用户id

    B=list(set(a1))

    # data_class = pd.DataFrame(A,lista)

    #创建商品类别字典

    a = np.arange(len(A))

    lista = list(a)

    dict_class = dict(zip(A,lista))

    print dict_class

    f=open('class.txt','w')

    for k ,v in dict_class.items():

    f.write(str(k)+'\t'+str(v)+'\n')

    f.close()

    #计算运行时间

    start=time.clock()

    #创建大字典存储数据

    dictall = {}

    for i in xrange(len(a1)):

    if a1[i] in dictall.keys():

    value = dictall[a1[i]]

    j = dict_class[a2[i]]

    value[j] = a3[i]

    dictall[a1[i]]=value

    else:

    value = list(np.zeros(len(A)))

    j = dict_class[a2[i]]

    value[j] = a3[i]

    dictall[a1[i]]=value

    #将字典转化为dataframe

    dictall1 = pd.DataFrame(dictall)

    dictall_matrix = dictall1.T

    print dictall_matrix

    end = time.clock()

    print "赋值过程运行时间是:%f s"%(end-start)

    输出结果:

    {'\xe4\xb8\x93\xe4\xb8\x9a\xe8\xbf\x90\xe5\x8a\xa8': 4, '\xe7\x94\xb7\xe5\xa3\xab\xe6\x89\x8b\xe8\xa2\x8b': 1, '\xe5\xa5\xb3\xe5\xa3\xab\xe6\x89\x8b\xe8\xa2\x8b': 2, '\xe7\xab\xa5\xe8\xa3\x85\xef\xbc\x88\xe4\xb8\xad\xe5\xa4\xa7\xe7\xab\xa5)': 3, '\xe7\x94\xb7\xe5\xa3\xab\xe9\x85\x8d\xe4\xbb\xb6': 9, '\xe5\x9f\xba\xe7\xa1\x80\xe5\x86\x85\xe8\xa1\xa3': 8, '\xe6\x97\xb6\xe5\xb0\x9a': 6, '\xe6\xb7\x91\xe5\xa5\xb3\xe8\xa3\x85': 7, '\xe5\x95\x86\xe5\x8a\xa1\xe6\xad\xa3\xe8\xa3\x85': 5, '\xe5\xa5\xb3\xe9\xa5\xb0\xe5\x93\x81': 0}

    1 2 3 4 5 6 7 8 9

    0 1882 0 0 0 0 0 0 0 0

    0 0 345 0 0 0 0 2491 0 0

    0 0 0 0 0 0 0 0 328 0

    86 0 0 0 0 4985 969 0 0 0

    0 0 0 2033 399 0 0 0 0 38

    赋值过程运行时间是:0.004497 s

    linux环境下字符编码不同,class.txt:

    专业运动 4

    男士手袋 1

    女士手袋 2

    童装(中大童) 3

    男士配件 9

    基础内衣 8

    时尚 6

    淑女装 7

    商务正装 5

    女饰品 0

    得到的dicta_matrix 就是我们拿来跑数据的格式,每一列是商品名称,每一行是用户id

    现在我们来跑AE模型(Auto-encoder),简单说说AE模型,主要步骤很简单,有三层,输入-隐含-输出,把数据input进去,encode然后再decode,cost_function就是output与input之间的“差值”(有公式),差值越小,目标函数值越优。简单地说,就是你输入n维的数据,输出的还是n维的数据,有人可能会问,这有什么用呢,其实也没什么用,主要是能够把数据缩放,如果你输入的维数比较大,譬如实际的特征是几千维的,全部拿到算法里跑,效果不见得好,因为并不是所有特征都是有用的,用AE模型后,你可以压缩成m维(就是隐含层的节点数),如果输出的数据和原始数据的大小变换比例差不多,就证明这个隐含层的数据是可用的。这样看来好像和降维的思想类似,当然AE模型的用法远不止于此,具体贴一篇梁博的博文

    不过梁博的博文是用c++写的,这里使用python写的代码(开源代码,有少量改动):

    #/usr/bin/python

    #coding:utf-8

    import pandas as pd

    import numpy as np

    import matplotlib.pyplot as plt

    from sklearn import preprocessing

    class AutoEncoder():

    """ Auto Encoder

    layer 1 2 ... ... L-1 L

    W 0 1 ... ... L-2

    B 0 1 ... ... L-2

    Z 0 1 ... L-3 L-2

    A 0 1 ... L-3 L-2

    """

    def __init__(self, X, Y, nNodes):

    # training samples

    self.X = X

    self.Y = Y

    # number of samples

    self.M = len(self.X)

    # layers of networks

    self.nLayers = len(nNodes)

    # nodes at layers

    self.nNodes = nNodes

    # parameters of networks

    self.W = list()

    self.B = list()

    self.dW = list()

    self.dB = list()

    self.A = list()

    self.Z = list()

    self.delta = list()

    for iLayer in range(self.nLayers - 1):

    self.W.append( np.random.rand(nNodes[iLayer]*nNodes[iLayer+1]).reshape(nNodes[iLayer],nNodes[iLayer+1]) )

    self.B.append( np.random.rand(nNodes[iLayer+1]) )

    self.dW.append( np.zeros([nNodes[iLayer], nNodes[iLayer+1]]) )

    self.dB.append( np.zeros(nNodes[iLayer+1]) )

    self.A.append( np.zeros(nNodes[iLayer+1]) )

    self.Z.append( np.zeros(nNodes[iLayer+1]) )

    self.delta.append( np.zeros(nNodes[iLayer+1]) )

    # value of cost function

    self.Jw = 0.0

    # active function (logistic function)

    self.sigmod = lambda z: 1.0 / (1.0 + np.exp(-z))

    # learning rate 1.2

    self.alpha = 2.5

    # steps of iteration 30000

    self.steps = 10000

    def BackPropAlgorithm(self):

    # clear values

    self.Jw -= self.Jw

    for iLayer in range(self.nLayers-1):

    self.dW[iLayer] -= self.dW[iLayer]

    self.dB[iLayer] -= self.dB[iLayer]

    # propagation (iteration over M samples)

    for i in range(self.M):

    # Forward propagation

    for iLayer in range(self.nLayers - 1):

    if iLayer==0: # first layer

    self.Z[iLayer] = np.dot(self.X[i], self.W[iLayer])

    else:

    self.Z[iLayer] = np.dot(self.A[iLayer-1], self.W[iLayer])

    self.A[iLayer] = self.sigmod(self.Z[iLayer] + self.B[iLayer])

    # Back propagation

    for iLayer in range(self.nLayers - 1)[::-1]: # reserve

    if iLayer==self.nLayers-2:# last layer

    self.delta[iLayer] = -(self.X[i] - self.A[iLayer]) * (self.A[iLayer]*(1-self.A[iLayer]))

    self.Jw += np.dot(self.Y[i] - self.A[iLayer], self.Y[i] - self.A[iLayer])/self.M

    else:

    self.delta[iLayer] = np.dot(self.W[iLayer].T, self.delta[iLayer+1]) * (self.A[iLayer]*(1-self.A[iLayer]))

    # calculate dW and dB

    if iLayer==0:

    self.dW[iLayer] += self.X[i][:, np.newaxis] * self.delta[iLayer][:, np.newaxis].T

    else:

    self.dW[iLayer] += self.A[iLayer-1][:, np.newaxis] * self.delta[iLayer][:, np.newaxis].T

    self.dB[iLayer] += self.delta[iLayer]

    # update

    for iLayer in range(self.nLayers-1):

    self.W[iLayer] -= (self.alpha/self.M)*self.dW[iLayer]

    self.B[iLayer] -= (self.alpha/self.M)*self.dB[iLayer]

    def PlainAutoEncoder(self):

    for i in range(self.steps):

    self.BackPropAlgorithm()

    print "step:%d" % i, "Jw=%f" % self.Jw

    def ValidateAutoEncoder(self):

    for i in range(self.M):

    print self.X[i]

    for iLayer in range(self.nLayers - 1):

    if iLayer==0: # input layer

    self.Z[iLayer] = np.dot(self.X[i], self.W[iLayer])

    else:

    self.Z[iLayer] = np.dot(self.A[iLayer-1], self.W[iLayer])

    self.A[iLayer] = self.sigmod(self.Z[iLayer] + self.B[iLayer])

    print "\t layer=%d" % iLayer, self.A[iLayer]

    data=[]

    index=[]

    f=open('./data_matrix.txt','r')

    for line in f.readlines():

    ss=line.replace('\n','').split('\t')

    index.append(ss[0])

    ss1=ss[1].split(' ')

    tmp=[]

    for i in xrange(len(ss1)):

    tmp.append(float(ss1[i]))

    data.append(tmp)

    f.close()

    x = np.array(data)

    #归一化处理

    xx = preprocessing.scale(x)

    nNodes = np.array([ 10, 5, 10])

    ae3 = AutoEncoder(xx,xx,nNodes)

    ae3.PlainAutoEncoder()

    ae3.ValidateAutoEncoder()

    #这是个例子,输出的结果也是这个

    # xx = np.array([[0,0,0,0,0,0,0,1], [0,0,0,0,0,0,1,0], [0,0,0,0,0,1,0,0], [0,0,0,0,1,0,0,0],[0,0,0,1,0,0,0,0], [0,0,1,0,0,0,0,0]])

    # nNodes = np.array([ 8, 3, 8 ])

    # ae2 = AutoEncoder(xx,xx,nNodes)

    # ae2.PlainAutoEncoder()

    # ae2.ValidateAutoEncoder()

    这里我拿的例子做的结果,真实数据在服务器上跑,大家看看这道啥意思就行了[0 0 0 0 0 0 0 1]

    layer=0 [ 0.76654705 0.04221051 0.01185895]

    layer=1 [ 4.67403977e-03 5.18624788e-03 2.03185410e-02 1.24383559e-02

    1.54423619e-02 1.69197292e-03 2.34471751e-05 9.72956513e-01]

    [0 0 0 0 0 0 1 0]

    layer=0 [ 0.08178768 0.96348458 0.98583155]

    layer=1 [ 8.18926274e-04 7.30041977e-04 1.06452565e-02 9.94423121e-03

    3.47329848e-03 1.32582980e-02 9.80648863e-01 8.42319408e-08]

    [0 0 0 0 0 1 0 0]

    layer=0 [ 0.04752084 0.01144966 0.67313608]

    layer=1 [ 4.38577163e-03 4.12704649e-03 1.83408905e-02 1.59209302e-05

    2.32400619e-02 9.71429772e-01 1.78538577e-02 2.20897151e-03]

    [0 0 0 0 1 0 0 0]

    layer=0 [ 0.00819346 0.37410028 0.0207633 ]

    layer=1 [ 8.17965283e-03 7.94760145e-03 4.59916741e-05 2.03558668e-02

    9.68811657e-01 2.09241369e-02 6.19909778e-03 1.51964053e-02]

    [0 0 0 1 0 0 0 0]

    layer=0 [ 0.88632868 0.9892662 0.07575306]

    layer=1 [ 1.15787916e-03 1.25924912e-03 3.72748604e-03 9.79510789e-01

    1.09439392e-02 7.81892291e-08 1.06705286e-02 1.77993321e-02]

    [0 0 1 0 0 0 0 0]

    layer=0 [ 0.9862938 0.2677048 0.97331042]

    layer=1 [ 6.03115828e-04 6.37411444e-04 9.75530999e-01 4.06825647e-04

    2.66386294e-07 1.27802666e-02 8.66599313e-03 1.06025228e-02]

    可以很明显看layer1和原始数据是对应的,所以我们可以把layer0作为降维后的新数据。

    最后在进行聚类,这个就比较简单了,用sklearn的包,就几行代码:

    # !/usr/bin/python

    # coding:utf-8

    # Author :Charlotte

    from matplotlib import pyplot

    import scipy as sp

    import numpy as np

    import matplotlib.pyplot as plt

    from sklearn.cluster import KMeans

    from scipy import sparse

    import pandas as pd

    import Pycluster as pc

    from sklearn import preprocessing

    from sklearn.preprocessing import StandardScaler

    from sklearn import metrics

    import pickle

    from sklearn.externals import joblib

    #加载数据

    data = pd.read_table('data_new.txt',header = None,sep = " ")

    x = data.ix[:,1:141]

    card = data.ix[:,0]

    x1 = np.array(x)

    xx = preprocessing.scale(x1)

    num_clusters = 5

    clf = KMeans(n_clusters=num_clusters, n_init=1, n_jobs = -1,verbose=1)

    clf.fit(xx)

    print(clf.labels_)

    labels = clf.labels_

    #score是轮廓系数

    score = metrics.silhouette_score(xx, labels)

    # clf.inertia_用来评估簇的个数是否合适,距离越小说明簇分的越好

    print clf.inertia_

    print score

    这个数据是拿来做例子的,维度少,效果不明显,真实环境下的数据是30W*142维的,写的mapreduce程序进行数据处理,然后通过AE模型降到50维后,两者的clf.inertia_和silhouette(轮廓系数)有显著差异:

    所以可以看到没有用AE模型直接聚类的模型跑完后的clf.inertia_比用了AE模型之后跑完的clf.inertia_大了几个数量级,AE的效果还是很显著的。

    以上是随手整理的,如有错误,欢迎指正:)

    老师介绍:胡晓曼老师(Charlotte),高级算法工程师 ,博客专家;

    擅长用通俗易懂的方式讲解深度学习和机器学习算法,熟悉Tensorflow,PaddlePaddle等深度学习框架,负责过多个机器学习落地项目,如垃圾评论自动过滤,用户分级精准营销,分布式深度学习平台搭建等,都取了的不错的效果。

    出处:https://www.hellobi.com/u/CharlotteDataMining/articles

    三个月教你从零入门人工智能!| 深度学习精华实践课程

    https://edu.hellobi.com/course/268

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