解读python如何实现决策树算法

数据描述

每条数据项储存在列表中，最后一列储存结果

多条数据项形成数据集

data=[[d1,d2,d3...dn,result],   [d1,d2,d3...dn,result],        .        .   [d1,d2,d3...dn,result]]

决策树数据结构

class DecisionNode:  '''决策树节点  '''     def __init__(self,col=-1,value=None,results=None,tb=None,fb=None):    '''初始化决策树节点         args：        col -- 按数据集的col列划分数据集    value -- 以value作为划分col列的参照    result -- 只有叶子节点有，代表最终划分出的子数据集结果统计信息。｛‘结果'：结果出现次数｝    rb,fb -- 代表左右子树    '''    self.col=col    self.value=value    self.results=results    self.tb=tb    self.fb=fb

决策树分类的最终结果是将数据项划分出了若干子集，其中每个子集的结果都一样，所以这里采用｛‘结果'：结果出现次数｝的方式表达每个子集

def pideset(rows,column,value):  '''依据数据集rows的column列的值，判断其与参考值value的关系对数据集进行拆分    返回两个数据集  '''  split_function=None  #value是数值类型  if isinstance(value,int) or isinstance(value,float):    #定义lambda函数当row[column]>=value时返回true    split_function=lambda row:row[column]>=value  #value是字符类型  else:    #定义lambda函数当row[column]==value时返回true    split_function=lambda row:row[column]==value  #将数据集拆分成两个  set1=[row for row in rows if split_function(row)]  set2=[row for row in rows if not split_function(row)]  #返回两个数据集  return (set1,set2) def uniquecounts(rows):  '''计算数据集rows中有几种最终结果，计算结果出现次数，返回一个字典  '''  results={}  for row in rows:    r=row[len(row)-1]    if r not in results: results[r]=0    results[r]+=1  return results def giniimpurity(rows):  '''返回rows数据集的基尼不纯度  '''  total=len(rows)  counts=uniquecounts(rows)  imp=0  for k1 in counts:    p1=float(counts[k1])/total    for k2 in counts:      if k1==k2: continue      p2=float(counts[k2])/total      imp+=p1*p2  return imp def entropy(rows):  '''返回rows数据集的熵  '''  from math import log  log2=lambda x:log(x)/log(2)   results=uniquecounts(rows)  ent=0.0  for r in results.keys():    p=float(results[r])/len(rows)    ent=ent-p*log2(p)  return ent def build_tree(rows,scoref=entropy):  '''构造决策树  '''  if len(rows)==0: return DecisionNode()  current_score=scoref(rows)   # 最佳信息增益  best_gain=0.0  #  best_criteria=None  #最佳划分  best_sets=None   column_count=len(rows[0])-1  #遍历数据集的列，确定分割顺序  for col in range(0,column_count):    column_values={}    # 构造字典    for row in rows:      column_values[row[col]]=1    for value in column_values.keys():      (set1,set2)=pideset(rows,col,value)      p=float(len(set1))/len(rows)      # 计算信息增益      gain=current_score-p*scoref(set1)-(1-p)*scoref(set2)      if gain>best_gain and len(set1)>0 and len(set2)>0:        best_gain=gain        best_criteria=(col,value)        best_sets=(set1,set2)  # 如果划分的两个数据集熵小于原数据集，进一步划分它们  if best_gain>0:    trueBranch=build_tree(best_sets[0])    falseBranch=build_tree(best_sets[1])    return DecisionNode(col=best_criteria[0],value=best_criteria[1],            tb=trueBranch,fb=falseBranch)  # 如果划分的两个数据集熵不小于原数据集，停止划分  else:    return DecisionNode(results=uniquecounts(rows)) def print_tree(tree,indent=''):  if tree.results!=None:    print(str(tree.results))  else:    print(str(tree.col)+':'+str(tree.value)+'? ')    print(indent+'T->',end='')    print_tree(tree.tb,indent+' ')    print(indent+'F->',end='')    print_tree(tree.fb,indent+' ')  def getwidth(tree):  if tree.tb==None and tree.fb==None: return 1  return getwidth(tree.tb)+getwidth(tree.fb) def getdepth(tree):  if tree.tb==None and tree.fb==None: return 0  return max(getdepth(tree.tb),getdepth(tree.fb))+1  def drawtree(tree,jpeg='tree.jpg'):  w=getwidth(tree)*100  h=getdepth(tree)*100+120   img=Image.new('RGB',(w,h),(255,255,255))  draw=ImageDraw.Draw(img)   drawnode(draw,tree,w/2,20)  img.save(jpeg,'JPEG') def drawnode(draw,tree,x,y):  if tree.results==None:    # Get the width of each branch    w1=getwidth(tree.fb)*100    w2=getwidth(tree.tb)*100     # Determine the total space required by this node    left=x-(w1+w2)/2    right=x+(w1+w2)/2     # Draw the condition string    draw.text((x-20,y-10),str(tree.col)+':'+str(tree.value),(0,0,0))     # Draw links to the branches    draw.line((x,y,left+w1/2,y+100),fill=(255,0,0))    draw.line((x,y,right-w2/2,y+100),fill=(255,0,0))       # Draw the branch nodes    drawnode(draw,tree.fb,left+w1/2,y+100)    drawnode(draw,tree.tb,right-w2/2,y+100)  else:    txt=' /n'.join(['%s:%d'%v for v in tree.results.items()])    draw.text((x-20,y),txt,(0,0,0))