“unstack” a pandas column containing lists into multiple rows [duplicate]

Answer 1

UPDATE: generic vectorized approach - will work also for multiple columns DFs:

assuming we have the following DF:

In [159]: df
Out[159]:
   a          b  c
0  1     [1, 2]  5
1  2  [2, 3, 4]  6
2  3        [5]  7

Solution:

In [160]: lst_col = 'b'

In [161]: pd.DataFrame({
     ...:     col:np.repeat(df[col].values, df[lst_col].str.len())
     ...:     for col in df.columns.difference([lst_col])
     ...: }).assign(**{lst_col:np.concatenate(df[lst_col].values)})[df.columns.tolist()]
     ...:
Out[161]:
   a  b  c
0  1  1  5
1  1  2  5
2  2  2  6
3  2  3  6
4  2  4  6
5  3  5  7

Setup:

df = pd.DataFrame({
    "a" : [1,2,3],
    "b" : [[1,2],[2,3,4],[5]],
    "c" : [5,6,7]
})

Vectorized NumPy approach:

In [124]: pd.DataFrame({'a':np.repeat(df.a.values, df.b.str.len()),
                        'b':np.concatenate(df.b.values)})
Out[124]:
   a  b
0  1  1
1  1  2
2  2  2
3  2  3
4  2  4
5  3  5

OLD answer:

Try this:

In [89]: df.set_index('a', append=True).b.apply(pd.Series).stack().reset_index(level=[0, 2], drop=True).reset_index()
Out[89]:
   a    0
0  1  1.0
1  1  2.0
2  2  2.0
3  2  3.0
4  2  4.0
5  3  5.0

Or bit nicer solution provided by @Boud:

In [110]: df.set_index('a').b.apply(pd.Series).stack().reset_index(level=-1, drop=True).astype(int).reset_index()
Out[110]:
   a  0
0  1  1
1  1  2
2  2  2
3  2  3
4  2  4
5  3  5

Answer 2

Here is another approach with itertuples -

df = pd.DataFrame({"a" : [1,2,3], "b" : [[1,2],[2,3,4],[5]]})

data = []

for i in df.itertuples():
    lst = i[2]
    for col2 in lst:
        data.append([i[1], col2])

df_output = pd.DataFrame(data =data, columns=df.columns)
df_output 

Output is -

        a   b
    0   1   1
    1   1   2
    2   2   2
    3   2   3
    4   2   4
    5   3   5

Edit: You can also compress the loops into a single code and populate data as -

data = [[i[1], col2] for i in df.itertuples() for col2 in i[2]]