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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