問題描述

最近遇到一個正則表達式替換的問題

time數(shù)據里面的每條數(shù)據前面都有[0]= [1]= [2]= [3]=這個索引：

['time']={[0]={['status']=true,['ac']=1,['bg']=2},[1]={['status']=true,['ac']=1,['bg']=2},[2]={['status']=true,['ac']=1,['bg']=2},}

因為一些原因前面的索引沒了，只能用正則來加上，問題是time里面的數(shù)據數(shù)量是不一樣的

['time']={{['status']=true,['ac']=1,['bg']=2},}['time']={{['status']=true,['ac']=1,['bg']=2},{['status']=true,['ac']=1,['bg']=2},}['time']={{['status']=true,['ac']=1,['bg']=2},{['status']=true,['ac']=1,['bg']=2},{['status']=true,['ac']=1,['bg']=2},}

有沒有方法自動在前面加順序的[0]= [1]= [2]= [3]=

補充：

錯誤的數(shù)據是在一起的，而且time里面的數(shù)據順序不相同，如下：

['time1']={{['status']=true,['ac']=1,['bg']=2},},['time2']={{['status']=true,['ac']=1,['bg']=2},{['status']=true,['ac']=1,['bg']=2},},['time3']={{['status']=true,['ac']=1,['bg']=2},{['status']=true,['ac']=1,['bg']=2},{['status']=true,['ac']=1,['bg']=2},}

想改成：

['time1']={[0]={['status']=true,['ac']=1,['bg']=2},},['time2']={[0]={['status']=true,['ac']=1,['bg']=2},[1]={['status']=true,['ac']=1,['bg']=2},},['time3']={[0]={['status']=true,['ac']=1,['bg']=2},[1]={['status']=true,['ac']=1,['bg']=2},[2]={['status']=true,['ac']=1,['bg']=2},}

問題解答

>>> import re>>> s=’['time']={{['status']=true,['ac']=1,['bg']=2},{['status']=true,['ac']=1,['bg']=2},{['status']=true,['ac']=1,['bg']=2},}’>>> n=0>>> def repl(m): global n rslt=’[%d]=%s’%(n,m.group(0)) n+=1 return rslt>>> p=re.compile(r’{[^{}]+},’)>>> p.sub(repl,s)’['time']={[0]={['status']=true,['ac']=1,['bg']=2},[1]={['status']=true,['ac']=1,['bg']=2},[2]={['status']=true,['ac']=1,['bg']=2},}’回答2：

i = 0def func(x): global i s = ’[%d]=%s’ % (i,x) i += 1 return s import rea = ’['time']={{['status']=true,['ac']=1,['bg']=2},{['status']=true,['ac']=1,['bg']=2},}’print re.sub(’{['status'’,lambda m:func(m.group(0)),a)

寫的不好，見笑了