先收藏。。。。。。。。。。。。
本文为笔者在学习周志华老师的机器学习教材后，写的课后习题的的编程题。之前放在答案的博文中，现在重新进行整理，将需要实现代码的部分单独拿出来，慢慢积累。希望能写一个机器学习算法实现的系列。
本文主要包括：
1、logistics回归
2、python库：
numpymatplotlibpandas使用的数据集：机器学习教材上的西瓜数据集3.0α
logistic回归：参考《机器学习实战》的内容。本题分别写了梯度上升方法以及随机梯度上升方法。对书本上的程序做了一点点改动
# -*- coding: cp936 -*- from numpy import * import pandas as pd import matplotlib.pyplot as plt #读入csv文件数据 df=pd.read_csv('watermelon_3a.csv') m,n=shape(dataMat) df['norm']=ones((m,1)) dataMat=array(df[['norm','density','ratio_sugar']].values[:,:]) labelMat=mat(df['label'].values[:]).transpose() #sigmoid函数 def sigmoid(inX): return 1.0/(1+exp(-inX)) #梯度上升算法 def gradAscent(dataMat,labelMat): m,n=shape(df.values) alpha=0.1 maxCycles=500 weights=array(ones((n,1))) for k in range(maxCycles): a=dot(dataMat,weights) h=sigmoid(a) error=(labelMat-h) weights=weights+alpha*dot(dataMat.transpose(),error) return weights #随机梯度上升 def randomgradAscent(dataMat,label,numIter=50): m,n=shape(dataMat) weights=ones(n) for j in range(numIter): dataIndex=range(m) for i in range(m): alpha=40/(1.0+j+i)+0.2 randIndex_Index=int(random.uniform(0,len(dataIndex))) randIndex=dataIndex[randIndex_Index] h=sigmoid(sum(dot(dataMat[randIndex],weights))) error=(label[randIndex]-h) weights=weights+alpha*error[0,0]*(dataMat[randIndex].transpose()) del(dataIndex[randIndex_Index]) return weights #画图 def plotBestFit(weights): m=shape(dataMat)[0] xcord1=[] ycord1=[] xcord2=[] ycord2=[] for i in range(m): if labelMat[i]==1: xcord1.append(dataMat[i,1]) ycord1.append(dataMat[i,2]) else: xcord2.append(dataMat[i,1]) ycord2.append(dataMat[i,2]) plt.figure(1) ax=plt.subplot(111) ax.scatter(xcord1,ycord1,s=30,c='red',marker='s') ax.scatter(xcord2,ycord2,s=30,c='green') x=arange(0.2,0.8,0.1) y=array((-weights[0]-weights[1]*x)/weights[2]) print shape(x) print shape(y) plt.sca(ax) plt.plot(x,y)#ramdomgradAscent #plt.plot(x,y[0]) #gradAscent plt.xlabel('density') plt.ylabel('ratio_sugar') #plt.title('gradAscent logistic regression') plt.title('ramdom gradAscent logistic regression') plt.show() #weights=gradAscent(dataMat,labelMat) weights=randomgradAscent(dataMat,labelMat) plotBestFit(weights)
梯度上升法得到的结果如下：
随机梯度上升法得到的结果如下：
可以看出，两种方法的效果基本差不多。但是随机梯度上升方法所需要的迭代次数要少很多
LDA的编程主要参考书上P62的3.39 以及P61的3.33这两个式子。由于用公式可以直接算出，因此比较简单公式如下：
代码如下：
# -*- coding: cp936 -*- from numpy import * import numpy as np import pandas as pd import matplotlib.pyplot as plt df=pd.read_csv('watermelon_3a.csv') def calulate_w(): df1=df[df.label==1] df2=df[df.label==0] X1=df1.values[:,1:3] X0=df2.values[:,1:3] mean1=array([mean(X1[:,0]),mean(X1[:,1])]) mean0=array([mean(X0[:,0]),mean(X0[:,1])]) m1=shape(X1)[0] sw=zeros(shape=(2,2)) for i in range(m1): xsmean=mat(X1[i,:]-mean1) sw+=xsmean.transpose()*xsmean m0=shape(X0)[0] for i in range(m0): xsmean=mat(X0[i,:]-mean0) sw+=xsmean.transpose()*xsmean w=(mean0-mean1)*(mat(sw).I) return w def plot(w): dataMat=array(df[['density','ratio_sugar']].values[:,:]) labelMat=mat(df['label'].values[:]).transpose() m=shape(dataMat)[0] xcord1=[] ycord1=[] xcord2=[] ycord2=[] for i in range(m): if labelMat[i]==1: xcord1.append(dataMat[i,0]) ycord1.append(dataMat[i,1]) else: xcord2.append(dataMat[i,0]) ycord2.append(dataMat[i,1]) plt.figure(1) ax=plt.subplot(111) ax.scatter(xcord1,ycord1,s=30,c='red',marker='s') ax.scatter(xcord2,ycord2,s=30,c='green') x=arange(-0.2,0.8,0.1) y=array((-w[0,0]*x)/w[0,1]) print shape(x) print shape(y) plt.sca(ax) #plt.plot(x,y)#ramdomgradAscent plt.plot(x,y) #gradAscent plt.xlabel('density') plt.ylabel('ratio_sugar') plt.title('LDA') plt.show() w=calulate_w() plot(w)
结果如下：
对应的w值为：
[ -6.62487509e-04, -9.36728168e-01]
由于数据分布的关系，所以LDA的效果不太明显。所以我改了几个label=0的样例的数值，重新运行程序得到结果如下：
效果比较明显，对应的w值为：
[-0.60311161, -0.67601433]
转自：/c?m=9d78d513d9d430db4f9be0697b14c0101f4381132ba6d70209d6843890732f43506793ac57270772d7d20d1016db4d4bea81743971597deb8f8fc814d2e1d46e6d9f26476d01d61f4f860eafbc1764977c875a9ef34ea1a7b57accef8c959a49008a155e2bdea7960c57529934ae552ce4a59b49105a10bd&p=ce6fc64ad4d807f449bd9b7d0d1796&newp=c26ada15d9c041ae17a6c7710f0a88231610db2151dcd101298ffe0cc4241a1a1a3aecbf21261b01d4c67a6606a94c5de1f53373310434f1f689df08d2ecce7e60c3&user=baidu&fm=sc&query=%CF%DF%D0%D4%C5%D0%B1%F0%B7%D6%CE%F6+python&qid=ccbe92e80000a2cb&p1=1
