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))

对测试数据进行分类（附带处理缺失数据）

def mdclassify(observation,tree):
  '''对缺失数据进行分类

  args：
  observation -- 发生信息缺失的数据项
  tree -- 训练完成的决策树

  返回代表该分类的结果字典
  '''

  # 判断数据是否到达叶节点
  if tree.results!=None:
    # 已经到达叶节点，返回结果result
    return tree.results
  else:
    # 对数据项的col列进行分析
    v=observation[tree.col]

    # 若col列数据缺失
    if v==None:
      #对tree的左右子树分别使用mdclassify，tr是左子树得到的结果字典，fr是右子树得到的结果字典
      tr,fr=mdclassify(observation,tree.tb),mdclassify(observation,tree.fb)

      # 分别以结果占总数比例计算得到左右子树的权重
      tcount=sum(tr.values())
      fcount=sum(fr.values())
      tw=float(tcount)/(tcount+fcount)
      fw=float(fcount)/(tcount+fcount)
      result={}

      # 计算左右子树的加权平均
      for k,v in tr.items():
        result[k]=v*tw
      for k,v in fr.items():
        # fr的结果k有可能并不在tr中，在result中初始化k
        if k not in result:
          result[k]=0
        # fr的结果累加到result中
        result[k]+=v*fw
      return result

    # col列没有缺失，继续沿决策树分类
    else:
      if isinstance(v,int) or isinstance(v,float):
        if v>=tree.value: branch=tree.tb
        else: branch=tree.fb
      else:
        if v==tree.value: branch=tree.tb
        else: branch=tree.fb
      return mdclassify(observation,branch)

tree=build_tree(my_data)
print(mdclassify(['google',None,'yes',None],tree))
print(mdclassify(['google','France',None,None],tree))

决策树剪枝

def prune(tree,mingain):
  '''对决策树进行剪枝

  args：
  tree -- 决策树
  mingain -- 最小信息增益

  返回
  '''
  # 修剪非叶节点
  if tree.tb.results==None:
    prune(tree.tb,mingain)
  if tree.fb.results==None:
    prune(tree.fb,mingain)
  #合并两个叶子节点
  if tree.tb.results!=None and tree.fb.results!=None:
    tb,fb=[],[]
    for v,c in tree.tb.results.items():
      tb+=[[v]]*c
    for v,c in tree.fb.results.items():
      fb+=[[v]]*c
    #计算熵减少情况
    delta=entropy(tb+fb)-(entropy(tb)+entropy(fb)/2)
    #熵的增加量小于mingain，可以合并分支
    if delta<mingain:
      tree.tb,tree.fb=None,None
      tree.results=uniquecounts(tb+fb)