数据挖掘--决策树基本应用实例

决策树

http://mayuxiang.sinaapp.com/?p=155

http://www.engr.uvic.ca/~seng474/

#!/usr/bin/python#coding:utf-8my_data=[['slashdot','USA','yes',18,'None'],        ['google','France','yes',23,'Premium'],        ['digg','USA','yes',24,'Basic'],        ['kiwitobes','France','yes',23,'Basic'],        ['google','UK','no',21,'Premium'],        ['(direct)','New Zealand','no',12,'None'],        ['(direct)','UK','no',21,'Basic'],        ['google','USA','no',24,'Premium'],        ['slashdot','France','yes',19,'None'],        ['digg','USA','no',18,'None'],        ['google','UK','no',18,'None'],        ['kiwitobes','UK','no',19,'None'],        ['digg','New Zealand','yes',12,'Basic'],        ['slashdot','UK','no',21,'None'],        ['google','UK','yes',18,'Basic'],        ['kiwitobes','France','yes',19,'Basic']]class decisionnode:  def __init__(self,col=-1,value=None,results=None,tb=None,fb=None):    self.col=col    self.value=value    self.results=results    self.tb=tb    self.fb=fb    # 根据某一列进行子集划分，可同时处理数字和非数值两类数据def divideset(rows,column,value):    # 定义一个函数来计算出某一行是属于Group1 （值为True）还是Group2    #（值为false）    split_function=None    if isinstance(value,int) or isinstance(value,float):     # 如果变量为数值类型，那么用“大于等于”来判断        split_function=lambda row:row[column]>=value    else:        split_function=lambda row:row[column]==value    # 将结果存于两个集合数组，并返回    set1=[row for row in rows if split_function(row)]    set2=[row for row in rows if not split_function(row)]    return (set1,set2)#结果集中每个结果出现的次数，如None字段出现2次def uniquecounts(rows):    classcounts = {}    for row in rows:        # we care about the last column, the class column        theclass = row[len(row)-1]        if theclass not in classcounts: classcounts[theclass]=0        classcounts[theclass]+=1    return classcounts# 计算熵值 e=p(x)log(p(x))，熵值越低，划分效果越好def entropy(rows):    from math import log    log2=lambda x:log(x)/log(2)    results=uniquecounts(rows)  #结果集中每个结果出现的次数，    #开始计算熵值    ent=0.0    for r in results.keys():        p=float(results[r])/len(rows)        ent=ent-p*log2(p)    return ent# treepredict.divideset(treepredict.my_data,2,'yes')def buildtree(rows,scoref=entropy):    if len(rows)==0: return decisionnode()    # Set up some variables to track the best split 设置变量来捕获最优解    lowest_impurity = scoref(rows)    best_split = None    best_sets = None        column_count = len(rows[0])-1    for col in range(0,column_count):        # Generate the list of different values in this column        column_values = {}        for row in rows: column_values[row[col]] = 1        # Now divide the rows up for each value in this column        for value in column_values.keys():            (set1,set2) = divideset(rows,col,value)            exp_impurity = float(len(set1))/len(rows) * scoref(set1) + float(len(set2))/len(rows) * scoref(set2)            if exp_impurity < lowest_impurity and len(set1)>0 and len(set2)>0:                lowest_impurity = exp_impurity                best_split = (col,value)                best_sets = (set1,set2)        if lowest_impurity < scoref(rows):        trueBranch = buildtree(best_sets[0],scoref)        falseBranch = buildtree(best_sets[1],scoref)        return decisionnode(col=best_split[0],value=best_split[1],                            tb=trueBranch, fb=falseBranch)    else:        return decisionnode(results=uniquecounts(rows))def printtree(tree, indent='    '):    if tree.results != None:        print tree.results    else:        print str(tree.col) + ':' + str(tree.value) + '?'        print indent+'T->',        printtree(tree.tb, indent+'  ')        print indent+'F->',        printtree(tree.fb, indent+'  ')#print divideset(my_data,2,'yes')print uniquecounts(my_data)print entropy(my_data)tree = buildtree(my_data)printtree(tree)