将熊猫df中的句子拆分为单词行并为每个句子编号

如何解决将熊猫df中的句子拆分为单词行并为每个句子编号

我是python的新手，并且有一个如下所示的pandas数据框：

df =
sn  sent                                                    ent
0   ['an','apple','is','an','example','of','what?']   ['O','F','0','O','O']
1   ['a','potato','V','O']

我想创建另一个如下所示的熊猫数据框：

newdf= 
sn  sent        ent
0   an          O
    apple       F
    is          O
    an          O
    example     O
    of          O
    what?       O
1   a           O
    potato      V
    is          O
    an          O
    example     O
    of          O
    what?       O

我尝试了这段代码，最终得到了代码下方显示的内容

df.set_index('sn')
            .stack()
            .str.split(expand=True)
            .stack()
            .unstack(level=1)
            .reset_index(level=0,drop=0)

它接近我想要的，但似乎可以弄清楚其余部分

sn  sent        ent
0   ['an',['O',0   'apple',0   'is',0   'an',0   'example',0   'of',0   'what?',1   'a',1   'potato',1   'is',1   'an',1   'example',1   'of',1   'what?']    'O']

非常感谢任何指针

解决方法

df = pd.DataFrame({'sn': [0,1],'sent': [['an','apple','is','an','example','of','what?'],['a','potato','what?']],'ent': [['O','F','0','O','O'],['O','V','O']]})

df.apply(pd.Series.explode).set_index('sn')

结果：

       sent ent
sn             
0        an   O
0     apple   F
0        is   0
0        an   0
0   example   0
0        of   O
0     what?   O
1         a   O
1    potato   V
1        is   0
1        an   0
1   example   0
1        of   O
1     what?   O