如何通过标点符号拆分 pandas 列中的长字符串

Question

I have a df that looks like this:

words                                              col_a   col_b  
I guess, because I have thought over that. Um,       1       0 
That? yeah.                                          1       1
I don't always think you're up to something.         0       1                                                       

I want to split df.words wherever a punctuation character is present (.,?:;;) into a separate row. However I want to preserve the col_b and col_b values from the original row for each new row. For example, the above df should look like this:

words                                              col_a   col_b  
I guess,                                             1       0
because I have thought over that.                    1       0
Um,                                                  1       0 
That?                                                1       1
yeah.                                                1       1
I don't always think you're up to something.         0       1

Answer 1

One way is using str.findall with the pattern (.*?[.,?:;;]) to match any of these punctuation sings and the characters that preceed it (non greedy), and explode the resulting lists:

(df.assign(words=df.words.str.findall(r'(.*?[.,?!:;])'))
   .explode('words')
   .reset_index(drop=True))

                                          words  col_a  col_b
0                                      I guess,      1      0
1             because I have thought over that.      1      0
2                                           Um,      1      0
3                                         That?      1      1
4                                         yeah.      1      1
5  I don't always think you're up to something.      0      1