本节书摘来自异步社区出版社《写给程序员的数据挖掘实践指南》一书中的第5章，第5.4节，作者：【美】Ron Zacharski（扎哈尔斯基），更多章节内容可以访问云栖社区“异步社区”公众号查看。
回到上一章当中提到的来自卡内基梅隆大学的汽车MPG数据集，该数据集的格式如下：
下面试图基于气缸的数目、排水量（立方英寸）、功率、重量和加速时间预测汽车的MPG。我将所有392个实例放到mpgData.txt文件中，然后编写了如下的短Python程序，该程序利用分层采样方法将数据分到10个桶中（数据集及Python代码都可以从网站guidetodatamining.com下载）。
import random def buckets(filename, bucketName, separator, classColumn): """the original data is in the file named filename bucketName is the prefix for all the bucket names separator is the character that divides the columns (for ex., a tab or comma) and classColumn is the column that indicates the class""" # put the data in 10 buckets numberOfBuckets = 10 data = {} # first read in the data and divide by category with open(filename) as f: lines = f.readlines() for line in lines: if separator != '\t': line = line.replace(separator, '\t') # first get the category category = line.split()[classColumn] data.setdefault(category, []) data[category].append(line) # initialize the buckets buckets = [] for i in range(numberOfBuckets): buckets.append([]) # now for each category put the data into the buckets for k in data.keys(): #randomize order of instances for each class random.shuffle(data[k]) bNum = 0 # divide into buckets for item in data[k]: buckets[bNum].append(item) bNum = (bNum + 1) % numberOfBuckets # write to file for bNum in range(numberOfBuckets): f = open("%s-i" % (bucketName, bNum + 1), 'w') for item in buckets[bNum]: f.write(item) f.close() buckets("mpgData.txt", 'mpgData','\t',0)执行上述代码会产生10个分别为mpgData01、mpgData02… mpgData10的文件。
能否修改上一章中近邻算法的代码，以使test函数能够在刚刚构建的10个文件上进行10折交叉验证（该数据集可以从网站guidetodatamining.com下载）？
你的程序应该输出类似如下矩阵的混淆矩阵：
该解答只涉及如下方面：
修改initializer方法以便从9个桶中读取数据；
加入一个新的方法对一个桶中的数据进行测试；
加入一个新的过程来执行10折交叉验证过程。
下面依次来考察上述修改。
initializer方法的签名看起来如下：
def __init__(self, bucketPrefix, testBucketNumber, dataFormat):每个桶的文件名类似于mpgData-01、mpgData-02，等等。这种情况下，bucketPrefix将是“mpgData”，而testBucketNumber是包含测试数据的桶。如果testBucketNumber为3，则分类器将会在桶1、2、4、5、6、7、8、9、10上进行训练。dataFormat是一个如何解释数据中每列的字符串，比如：
"class num num num num num comment"它表示第一列代表实例的类别，下面5列代表实例的数值型属性，最后一列会被看成注释。
新的初始化方法的完整代码如下：
import copy class Classifier: def __init__(self, bucketPrefix, testBucketNumber, dataFormat): """ a classifier will be built from files with the bucketPrefix excluding the file with textBucketNumber. dataFormat is a string that describes how to interpret each line of the data files. For example, for the mpg data the format is: "class num num num num num comment" """ self.medianAndDeviation = [] # reading the data in from the file self.format = dataFormat.strip().split('\t') self.data = [] # for each of the buckets numbered 1 through 10: for i in range(1, 11): # if it is not the bucket we should ignore, read the data if i != testBucketNumber: filename = "%s-i" % (bucketPrefix, i) f = open(filename) lines = f.readlines() f.close() for line in lines: fields = line.strip().split('\t') ignore = [] vector = [] for i in range(len(fields)): if self.format[i] == 'num': vector.append(float(fields[i])) elif self.format[i] == 'comment': ignore.append(fields[i]) elif self.format[i] == 'class': classification = fields[i] self.data.append((classification, vector, ignore)) self.rawData = copy.deepcopy(self.data) # get length of instance vector self.vlen = len(self.data[0][1]) # now normalize the data for i in range(self.vlen): self.normalizeColumn(i)testBucket方法下面编写一个新的方法来测试一个桶中的数据。
def testBucket(self, bucketPrefix, bucketNumber): """Evaluate the classifier with data from the file bucketPrefix-bucketNumber""" filename = "%s-i" % (bucketPrefix, bucketNumber) f = open(filename) lines = f.readlines() totals = {} f.close() for line in lines: data = line.strip().split('\t') vector = [] classInColumn = -1 for i in range(len(self.format)): if self.format[i] == 'num': vector.append(float(data[i])) elif self.format[i] == 'class': classInColumn = i theRealClass = data[classInColumn] classifiedAs = self.classify(vector) totals.setdefault(theRealClass, {}) totals[theRealClass].setdefault(classifiedAs, 0) totals[theRealClass][classifiedAs] += 1 return totals它以bucketPrefix和bucketNumber为输入，如果前者为“mpgData”、后者为3的话，测试数据将会从文件mpgData-03中读取，而testBucket将会返回如下格式的字典：
{'35': {'35': 1, '20': 1, '30': 1}, '40': {'30': 1}, '30': {'35': 3, '30': 1, '45': 1, '25': 1}, '15': {'20': 3, '15': 4, '10': 1}, '10': {'15': 1}, '20': {'15': 2, '20': 4, '30': 2, '25': 1}, '25': {'30': 5, '25': 3}}字典的键代表的是实例的真实类别。例如，上面第一行表示真实类别为35mpg的实例的结果。每个键的值是另一部字典，该字典代表分类器对实例进行分类的结果。例如行
'15': {'20': 3, '15': 4, '10': 1},表示实际为15mpg的3个实例被错分到20mpg类别中，而有4个实例被正确分到15mpg中，1个实例被错分到10mpg中。
10折交叉验证的执行流程最后，我们需要编写一个过程来实现10折交叉验证。也就是说，我们要构造10个分类器。每个分类器利用9个桶中的数据进行训练，而将其余数据用于测试。
def tenfold(bucketPrefix, dataFormat): results = {} for i in range(1, 11): c = Classifier(bucketPrefix, i, dataFormat) t = c.testBucket(bucketPrefix, i) for (key, value) in t.items(): results.setdefault(key, {}) for (ckey, cvalue) in value.items(): results[key].setdefault(ckey, 0) results[key][ckey] += cvalue # now print results categories = list(results.keys()) categories.sort() print( "\n Classified as: ") header = " " subheader = " +" for category in categories: header += category + " " subheader += "----+" print (header) print (subheader) total = 0.0 correct = 0.0 for category in categories: row = category + " |" for c2 in categories: if c2 in results[category]: count = results[category][c2] else: count = 0 row += " %2i |" % count total += count if c2 == category: correct += count print(row) print(subheader) print("\n%5.3f percent correct" %((correct * 100) / total)) print("total of %i instances" % total) tenfold("mpgData", "class num num num num num comment")运行上述程序会产生如下结果：
相关资源：敏捷开发V1.0.pptx