kNN算法python代码学习

参考书籍：《机器学习实战》

题目：约会网站

# -*- coding: utf8 -*-from numpy import *import operatordef createDataSet():    group = array([[1.0,1.1], [1.0,1.0], [0,0], [0,0.1]])    labels = ['A','A','B','B']    return group, labelsdef classify0(inX, dataSet, labels, k):    #inX是用于分类的输入向量，dataSet是输入的训练样本集，标签向量为labels，参数k表示用于选择最近邻的数目    dataSetSize = dataSet.shape[0]  #得到数组的行数，.shape[1]是得到数组的列数    diffMat = tile(inX, (dataSetSize,1)) - dataSet #tile将原来的数组扩展成了四个一样的数组                                                   #tile(inX,(a,b))a代表将原数组高维度扩充（纵向），                                                   #b代表低维度扩充（横向），先b后a                                                   #diffMat得到了目标与训练数值之间的差值    sqDiffMat = diffMat**2    sqDistances = sqDiffMat.sum(axis=1) #每行元素相加（横向），结果转置为行；若axis=0代表每列相加    distances = sqDistances**0.5    sortedDistIndicies = distances.argsort()#升序排列    #选择距离最小的k个点    classCount = {}    for i in range(k):        voteIlabel = labels[sortedDistIndicies[i]]        classCount[voteIlabel] = classCount.get(voteIlabel,0)+1 #字典赋值，dict.get(key,0),key为字典中要查找的键，                                                                #如果值不在字典中返回0，default是None    #排序    sortedClassCount=sorted(classCount.iteritems(),key=operator.itemgetter(1),reverse=True)                                                                #sorted(iterable, cmp=None, key=None, reverse=False)                                                                #iterable是可迭代类型(如列表)，dict.iteritems()产生字典的迭代器                                                                #operator.itemgetter(1)用于获取第1个域的值（从0 开始）                                                                #key还可以用lambda表达式来构造                                                                #reverse = True代表降序,False代表默认升序    return sortedClassCount[0][0]    

为了将文件里的内容转化为矩阵：

def file2matrix(filename):    fr = open(filename)    arrayOLines = fr.readlines()    numberOfLines = len(arrayOLines)  #得到文件的行数    returnMat = zeros((numberOfLines, 3))  #为了简化处理，将矩阵的另一维长度设置为3，根据需求决定    classLabelVector = []    index = 0    for line in arrayOLines:        line = line.strip()  #去掉所有的回车符        listFromLine = line.split('\t')  #将字符串数据用‘\t’分隔成元素列表        returnMat[index,:] = listFromLine[0:3]  #将前三个元素作为特征        classLabelVector.append(int(listFromLine[-1]))  #可以使用-1索引列表最后一列元素                                                        #如果不告诉解释器为整型，它会将元素当作字符串        index += 1    return returnMat,classLabelVector

用test.py文件进行可视化测试：

import kNNimport matplotlibimport matplotlib.pyplot as pltfrom numpy import arraydef myfind(x,data):    return [a for a in range(len(data)) if data[a]==x]    datingDataMat,datingLabels = kNN.file2matrix('datingTestSet2.txt')fig = plt.figure()ax = fig.add_subplot(111)set1 = myfind(1,datingLabels)set2 = myfind(2,datingLabels)set3 = myfind(3,datingLabels)plt.plot( [datingDataMat[i,0] for i in set1],[datingDataMat[i,1] for i in set1], 'ro',label = '1') plt.plot( [datingDataMat[i,0] for i in set2],[datingDataMat[i,1] for i in set2],'bo', label = '2')plt.plot( [datingDataMat[i,0] for i in set3],[datingDataMat[i,1] for i in set3],'ko', label = '3')#为了解决legend的问题！！！！ax.set_xlabel('flying km per year')ax.set_ylabel('the percentage of playing games')plt.legend(loc = 'best')plt.show()

得到的结果为（只利用了三个特征中的两个，分别为每年获得的飞行常客里程数和玩视频游戏所占的百分比）：

为了避免绝对数值大的特征对整体样本间距的影响过大（如飞行里程数），所以将特征数值归一化：

def autoNorm(dataSet):    minVals = dataSet.min(0)#0可以使得从列中选择最小值，而不是当前行的最小值    maxVals = dataSet.max(0)    ranges = maxVals - minVals    normDataSet = zeros(shape(dataSet))    m = dataSet.shape[0]    normDataSet = dataSet - tile(minVals, (m,1))    normDataSet = normDataSet/tile(ranges, (m,1)) #newValue = (oldValue-min)/(max-min)    return normDataSet, ranges, minVals

接下来，我们需要一个函数对分类器的正确率进行测试。这里构造一个计数器变量，每当对数据进行一次错误分类时，变量的值就加1，程序执行完后计数器的结果除以数据点总数就是错误率。

def datingClassTest():    hoRatio = 0.20#选取的测试样本占总样本的百分比    datingDataMat, datingLabels = file2matrix('datingTestSet2.txt')    normMat, ranges, minVals = autoNorm(datingDataMat)    m = normMat.shape[0]    numTestVecs = int(m*hoRatio)#测试样本的数量，其他的样本作为训练样本    errorCount = 0.0    for i in range(numTestVecs):        classifierResult = classify0(normMat[i,:],normMat[numTestVecs:m,:],datingLabels[numTestVecs:m],3)        #各参数的含义见classify0函数定义。我们选取了前numTestVecs个样本作为测试样本，其余作为训练样本        print 'the classifier came back with: %d, the real answer is: %d'%(classifierResult,datingLabels[i])        if classifierResult != datingLabels[i]: errorCount += 1.0    print 'the total error rate is: %f' % (errorCount/float(numTestVecs))

利用此函数测试一下：

>>> reload(kNN)<module 'kNN' from 'C:\Users\mrzhang\Desktop\prac\python\kNN.py'>>>> kNN.datingClassTest()

得到的结果为：

the classifier came back with: 3, the real answer is: 3the classifier came back with: 2, the real answer is: 2the classifier came back with: 1, the real answer is: 1the classifier came back with: 1, the real answer is: 1the classifier came back with: 1, the real answer is: 1......省略一部分数据the classifier came back with: 3, the real answer is: 3the classifier came back with: 2, the real answer is: 2the classifier came back with: 1, the real answer is: 1the classifier came back with: 3, the real answer is: 1the total error rate is: 0.050000

错误率为5%, 通过改变类别数目和测试样本数目，这个数字会改变。
接下来，我们可以利用此分类器，通过输入未知对象的属性信息，由分类软件来判定Helen和某一对象的可交往程度：讨厌、一般喜欢、非常喜欢。

对未知对象的预测函数为：

def classifyPerson():    resultList = ['not at all','in small doses', 'in large doses']    percentTats = float( raw_input("persentage of time spent playing video games?"))    ffMiles = float( raw_input('frequent flier miles earned peryear?'))    iceCream = float(raw_input('liters of ice cream consumed per year?'))    datingDataMat, datingLabels = file2matrix('datingTestSet2.txt')    normMat, ranges, minVals = autoNorm(datingDataMat)    inArr = array([ffMiles, percentTats, iceCream])    classifierResult = classify0( (inArr-minVals)/ranges, normMat, datingLabels, 3)    print 'You will probably like this person:',resultList[classifierResult - 1]

我们来试验一下：

>>> kNN.classifyPerson()persentage of time spent playing video games?20frequent flier miles earned peryear?5000liters of ice cream consumed per year?1You will probably like this person: in large doses