python 正则表达式替换

问题描述

最近遇到一个正则表达式替换的问题

time数据里面的每条数据前面都有[0]= [1]= [2]= [3]=这个索引：

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

因为一些原因前面的索引没了，只能用正则来加上，问题是time里面的数据数量是不一样的

['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},}

有没有方法自动在前面加顺序的[0]= [1]= [2]= [3]=

补充：

错误的数据是在一起的，而且time里面的数据顺序不相同，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},}

问题解答

>>> 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)

写的不好，见笑了