解讀python如何實(shí)現(xiàn)決策樹算法
數(shù)據(jù)描述
每條數(shù)據(jù)項(xiàng)儲(chǔ)存在列表中,最后一列儲(chǔ)存結(jié)果
多條數(shù)據(jù)項(xiàng)形成數(shù)據(jù)集
data=[[d1,d2,d3...dn,result], [d1,d2,d3...dn,result], . . [d1,d2,d3...dn,result]]
決策樹數(shù)據(jù)結(jié)構(gòu)
class DecisionNode: '''決策樹節(jié)點(diǎn) ''' def __init__(self,col=-1,value=None,results=None,tb=None,fb=None): '''初始化決策樹節(jié)點(diǎn) args: col -- 按數(shù)據(jù)集的col列劃分?jǐn)?shù)據(jù)集 value -- 以value作為劃分col列的參照 result -- 只有葉子節(jié)點(diǎn)有,代表最終劃分出的子數(shù)據(jù)集結(jié)果統(tǒng)計(jì)信息。{‘結(jié)果':結(jié)果出現(xiàn)次數(shù)} rb,fb -- 代表左右子樹 ''' self.col=col self.value=value self.results=results self.tb=tb self.fb=fb
決策樹分類的最終結(jié)果是將數(shù)據(jù)項(xiàng)劃分出了若干子集,其中每個(gè)子集的結(jié)果都一樣,所以這里采用{‘結(jié)果':結(jié)果出現(xiàn)次數(shù)}的方式表達(dá)每個(gè)子集
def pideset(rows,column,value): '''依據(jù)數(shù)據(jù)集rows的column列的值,判斷其與參考值value的關(guān)系對(duì)數(shù)據(jù)集進(jìn)行拆分 返回兩個(gè)數(shù)據(jù)集 ''' split_function=None #value是數(shù)值類型 if isinstance(value,int) or isinstance(value,float): #定義lambda函數(shù)當(dāng)row[column]>=value時(shí)返回true split_function=lambda row:row[column]>=value #value是字符類型 else: #定義lambda函數(shù)當(dāng)row[column]==value時(shí)返回true split_function=lambda row:row[column]==value #將數(shù)據(jù)集拆分成兩個(gè) set1=[row for row in rows if split_function(row)] set2=[row for row in rows if not split_function(row)] #返回兩個(gè)數(shù)據(jù)集 return (set1,set2) def uniquecounts(rows): '''計(jì)算數(shù)據(jù)集rows中有幾種最終結(jié)果,計(jì)算結(jié)果出現(xiàn)次數(shù),返回一個(gè)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數(shù)據(jù)集的基尼不純度 ''' 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數(shù)據(jù)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): '''構(gòu)造決策樹 ''' if len(rows)==0: return DecisionNode() current_score=scoref(rows) # 最佳信息增益 best_gain=0.0 # best_criteria=None #最佳劃分 best_sets=None column_count=len(rows[0])-1 #遍歷數(shù)據(jù)集的列,確定分割順序 for col in range(0,column_count): column_values={} # 構(gòu)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) # 計(jì)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) # 如果劃分的兩個(gè)數(shù)據(jù)集熵小于原數(shù)據(jù)集,進(jìn)一步劃分它們 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) # 如果劃分的兩個(gè)數(shù)據(jù)集熵不小于原數(shù)據(jù)集,停止劃分 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))
對(duì)測(cè)試數(shù)據(jù)進(jìn)行分類(附帶處理缺失數(shù)據(jù))
def mdclassify(observation,tree): '''對(duì)缺失數(shù)據(jù)進(jìn)行分類 args: observation -- 發(fā)生信息缺失的數(shù)據(jù)項(xiàng) tree -- 訓(xùn)練完成的決策樹 返回代表該分類的結(jié)果字典 ''' # 判斷數(shù)據(jù)是否到達(dá)葉節(jié)點(diǎn) if tree.results!=None: # 已經(jīng)到達(dá)葉節(jié)點(diǎn),返回結(jié)果result return tree.results else: # 對(duì)數(shù)據(jù)項(xiàng)的col列進(jìn)行分析 v=observation[tree.col] # 若col列數(shù)據(jù)缺失 if v==None: #對(duì)tree的左右子樹分別使用mdclassify,tr是左子樹得到的結(jié)果字典,fr是右子樹得到的結(jié)果字典 tr,fr=mdclassify(observation,tree.tb),mdclassify(observation,tree.fb) # 分別以結(jié)果占總數(shù)比例計(jì)算得到左右子樹的權(quán)重 tcount=sum(tr.values()) fcount=sum(fr.values()) tw=float(tcount)/(tcount+fcount) fw=float(fcount)/(tcount+fcount) result={} # 計(jì)算左右子樹的加權(quán)平均 for k,v in tr.items(): result[k]=v*tw for k,v in fr.items(): # fr的結(jié)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的結(jié)果累加到result中 result[k]+=v*fw return result # col列沒有缺失,繼續(xù)沿決策樹分類 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))
決策樹剪枝
def prune(tree,mingain): '''對(duì)決策樹進(jìn)行剪枝 args: tree -- 決策樹 mingain -- 最小信息增益 返回 ''' # 修剪非葉節(jié)點(diǎn) if tree.tb.results==None: prune(tree.tb,mingain) if tree.fb.results==None: prune(tree.fb,mingain) #合并兩個(gè)葉子節(jié)點(diǎn) 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 #計(jì)算熵減少情況 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)
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