如何使用将每个元素从嵌套列表转换为数据框？

Question

my_df = pd.DataFrame({'ID':['12345','23456','34567'],
         'Info':[[['Rob Kardashian', '00052369', '1987-03-17', 'Reality Star'], ['Brooke Barry', '00213658', '2001-03-30', 'TikTok Star']],
                [['Bae De Leon', '00896351', '1997-08-02', 'Volleyball Player'],['Jonas Blue', '02369785', '1990-08-02', 'Music Producer'],['Albert Einstein', '65231478', '1879-03-14','Scientist']],
                [['Robert Downey Jr', '23897410', '1965-04-04', 'Actor'],['Stan Lee','35239856','1922-12-28','Publisher & Producer']]]})

大家好，我有上面的数据框，想将列 'Info' 中的元素转换为行。我试过了

[[pd.DataFrame(i) for i in k] for k in my_df ['Info'].tolist()]

但是输出不是我预期的。

预期产出：

在此先感谢您的帮助！

Answer 1

这是你想要的吗：

my_df = my_df.set_index('ID')
pd.DataFrame(np.concatenate(my_df.Info), \
             index=my_df.index.repeat(my_df.Info.str.len()))

Out[1102]:
                      0         1           2                     3
ID
12345    Rob Kardashian  00052369  1987-03-17          Reality Star
12345      Brooke Barry  00213658  2001-03-30           TikTok Star
23456       Bae De Leon  00896351  1997-08-02     Volleyball Player
23456        Jonas Blue  02369785  1990-08-02        Music Producer
23456   Albert Einstein  65231478  1879-03-14             Scientist
34567  Robert Downey Jr  23897410  1965-04-04                 Actor
34567          Stan Lee  35239856  1922-12-28  Publisher & Producer

注意：我将ID作为输出df的索引。如果您需要它作为一列，请按如下方式链接额外的 .reset_index：

pd.DataFrame(np.concatenate(my_df.Info), \
            index=my_df.index.repeat(my_df.Info.str.len())).reset_index()

Out[1106]:
      ID                 0         1           2                     3
0  12345    Rob Kardashian  00052369  1987-03-17          Reality Star
1  12345      Brooke Barry  00213658  2001-03-30           TikTok Star
2  23456       Bae De Leon  00896351  1997-08-02     Volleyball Player
3  23456        Jonas Blue  02369785  1990-08-02        Music Producer
4  23456   Albert Einstein  65231478  1879-03-14             Scientist
5  34567  Robert Downey Jr  23897410  1965-04-04                 Actor
6  34567          Stan Lee  35239856  1922-12-28  Publisher & Producer

Answer 2

您可以使用分组：

my_df.groupby("ID").Info.apply(lambda g: pd.DataFrame(g.iloc[0]))

这会为您汇总返回的数据帧：

>>> my_df.groupby("ID").Info.apply(lambda g: pd.DataFrame(g.iloc[0]))
                        0         1           2                     3
ID
12345 0    Rob Kardashian  00052369  1987-03-17          Reality Star
      1      Brooke Barry  00213658  2001-03-30           TikTok Star
23456 0       Bae De Leon  00896351  1997-08-02     Volleyball Player
      1        Jonas Blue  02369785  1990-08-02        Music Producer
      2   Albert Einstein  65231478  1879-03-14             Scientist
34567 0  Robert Downey Jr  23897410  1965-04-04                 Actor
      1          Stan Lee  35239856  1922-12-28  Publisher & Producer

然后您可以选择重置索引并删除 level_1 列：

expanded = my_df.groupby("ID").Info.apply(lambda g: pd.DataFrame(g.iloc[0]))
expanded.reset_index().drop("level_1", axis=1)

这给了你

      ID                 0         1           2                     3
0  12345    Rob Kardashian  00052369  1987-03-17          Reality Star
1  12345      Brooke Barry  00213658  2001-03-30           TikTok Star
2  23456       Bae De Leon  00896351  1997-08-02     Volleyball Player
3  23456        Jonas Blue  02369785  1990-08-02        Music Producer
4  23456   Albert Einstein  65231478  1879-03-14             Scientist
5  34567  Robert Downey Jr  23897410  1965-04-04                 Actor
6  34567          Stan Lee  35239856  1922-12-28  Publisher & Producer

因为这使用 GroupBy.apply()，但是我不希望它那么快。

将 Andy 和我的版本封装到运行时间试验的函数中确实表明使用我的版本将是较慢的选择：

In [99]: def np_concat(df):
    ...:     df = df.set_index('ID')
    ...:     pd.DataFrame(np.concatenate(my_df.Info), index=my_df.index.repeat(my_df.Info.str.len()))
    ...:

In [100]: def groupby(df):
     ...:    df = df.groupby("ID").Info.apply(lambda g: pd.DataFrame(g.iloc[0]))
     ...:    df.reset_index().drop("level_1", axis=1)
     ...:

In [101]: %timeit np_concat(my_df)
1.08 ms ± 107 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

In [102]: %timeit groupby(my_df)
6.33 ms ± 394 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

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