kNN算法python代码学习2-手写识别系统

构造的系统能识别数字0到9，已经将图像转换为文本格式。

为了使用之前构造的分类器，要将图像格式化处理为一个向量。

def img2vector(filename):
returnVect = zeros((1,1024))
fr = open(filename)
for i in range(32): #图片大小为32*32
lineStr = fr.readline()
for j in range(32):
returnVect[0,32*i+j] = int(lineStr[j])
return returnVect

我们得到了分类器可以识别的数据格式，接下来就要将数据输入到分类器，对分类器的准确率进行测试：

def handwritingClassTest():
hwLabels = []
trainingFileList = listdir('trainingDigits')#使用此函数之前要将from os import listdir写在文件开头
m = len(trainingFileList)
trainingMat = zeros((m,1024))#每一行代表一个图片数据
for i in range(m):
fileNameStr = trainingFileList[i]
fileStr = fileNameStr.split('.')[0]
classNumStr = int(fileStr.split('_')[0]) #通过两次split获得分类数字（label）
hwLabels.append(classNumStr)
trainingMat[i,:] = img2vector('trainingDigits/%s'% fileNameStr)
testFileList = listdir('testDigits')
errorCount = 0.0
mTest = len(testFileList)
for i in range(mTest):
fileNameStr = testFileList[i]
fileStr = fileNameStr.split('.')[0]
classNumStr = int(fileStr.split('_')[0])
vectorUnderTest = img2vector('testDigits/%s' % fileNameStr)
classifierResult = classify0(vectorUnderTest, trainingMat, hwLabels, 3)#由于文件中的值都在0和1之间，所以不再需要归一化
print 'the classifier came back with: %d, the real answer is: %d'%(classifierResult,classNumStr)
if classifierResult != classNumStr: errorCount += 1.0
print '\nthe total number of errors is: %d' % errorCount
print '\nthe total error rate is: %f' % (errorCount/float(mTest))

命令行中输入：

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

测试结果为：

the classifier came back with: 0, the real answer is: 0
the classifier came back with: 0, the real answer is: 0
.....省略一些
the classifier came back with: 9, the real answer is: 9
the classifier came back with: 9, the real answer is: 9

the total number of errors is: 11

the total error rate is: 0.011628

得到的错误率为1.16%。

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