有一個數(shù)據(jù)表idfidtitle1-1python2-1ruby3-1php4-1lisp51flask61django71webpy82rails93zend106dblog這是一個欄目表,title是欄目名稱,id是欄目id,fid是欄目的父id.構(gòu)建一個多級欄目分類通過python查詢處理,會得到下面的一個元祖,t=((1,-1,'python'),(2,-1,'ruby'),(3,-1,'php'),(4,-1,'lisp'),(5,1,'flask'),(6,1,'django'),(7,1,'webpy'),(8,2,'rails'),(9,3,'zend'),(10,6,'dblog'))希望通過python處理,轉(zhuǎn)換成下面的list字典樹l=[{'id':1,'fid':-1,'title':'python','son':[{'id':5,'fid':1,'title':'flask',},{'id':6,'fid':1,'title':'django','son':[{'id':10,'fid':6,'title':'dblog',},]},{'id':7,'fid':1,'title':'webpy',},]},{'id':2,'fid':-1,'title':'ruby','son':[{'id':8,'fid':2,'title':'rails',},]},{'id':3,'fid':-1,'title':'php','son':[{'id':9,'fid':3,'title':'zend',},]},{'id':4,'fid':-1,'title':'lisp',}]也就是類似網(wǎng)站的目錄,父欄目包含子欄目.自己寫了好幾個,感覺效率不夠好,求大神更pythonic的方法在stackoverflow有人回答大概如下：frompprintimportpprintl=[]entries={}forid,fid,titleint:entries[id]=entry={'id':id,'fid':fid,'title':title}iffid==-1:l.append(entry)else:parent=entries[fid]parent.setdefault('son',[]).append(entry)pprint(l)