如何在数据框架中分割列值

Posted 2021-04-28

我有一个这样的数据框架

PK  Name Mobile  questions          
1   Jack 12345   [{'question':"how are you","Response":"Fine"},{"question":"whats your age","Response":"i am 19"}]

2   kim 102345   [{'question':"how are you","Response":"Not Fine"},{"question":"whats your age","Response":"i am 29"}]
3  jame  420 

我希望输出的df是这样的

PK    Name   Mobile Question 1     Response 1  Question 2        Response 2
1     Jack   12345   How are you   Fine        Whats your age    i am 19
2     Kim    102345  How are you   Not Fine    Whats your age    i am 29
3     jame   420     

答案

您可以使用 explode 来先在每个列表中每个元素创建一行。然后从这个爆炸的系列中创建一个数据框架，并保留索引。assign 一列来获取每行每索引组的增量值，然后 set_index 和 unstack 最后重命名列和 join 归原配

# create a row per element in each list in each row
s = df['questions'].explode()

# create the dataframe and reshape
df_qr = pd.DataFrame(s.tolist(), index=s.index)
          .assign(cc=lambda x: x.groupby(level=0).cumcount()+1)
          .set_index('cc', append=True).unstack()

#flatten columns names
df_qr.columns = [f'{col[0]} {col[1]}' for col in df_qr.columns]

# join back to df
df_f = df.drop('questions', axis=1).join(df_qr, how='left')

print (df_f)
   PK  Name  Mobile   question 1      question 2 Response 1 Response 2
0   1  Jack   12345  how are you  whats your age       Fine    i am 19
1   2   kim  102345  how are you  whats your age   Not Fine    i am 29

编辑，如果有些行是emppty字符串而不是lis，那么就可以创建 s 这样一来。

s = df.loc[df['questions'].apply(lambda x: isinstance(x, list)), 'questions'].explode()