reindex df on axis with duplicates

I have the following df:

df = pd.DataFrame({"col1": [0,0,0,1,1,1], "col2": ["a", "b", "c", "D", "E", "F"] }) 

and want to transform it in

df_new = pd.DataFrame({"col1": [0,1,0,1,0,1], "col2": ["a", "D", "b", "E", "c", "F"] })

How can I achieve this?

So far I have tried

df.reset_index()

resulting in

index  col1 col2
0      0     0    a
1      1     0    b
2      2     0    c
3      3     1    D
4      4     1    E
5      5     1    F

and

df_test.reindex()

which did not change the df at all.

Use a custom sorting with help of grouby.cumcount as key to sort_values and a stable sorting method:

df_new = df.sort_values(by='col1', key=lambda x: df.groupby(x).cumcount(),
                        kind='stable', ignore_index=True)

Or with numpy.argsort:

df_new = df.iloc[np.argsort(df.groupby('col1').cumcount())]

Output:

   col1 col2
0     0    a
1     1    D
2     0    b
3     1    E
4     0    c
5     1    F

This works by creating an intermediate sorter key that will be used to reindex your rows:

df.groupby('col1').cumcount()

0    0
1    1
2    2
3    0
4    1
5    2
dtype: int64

Use DataFrame.sort_values with counter by GroupBy.cumcount:

df = df.sort_values('col1', key=lambda x: df.groupby('col1').cumcount(), ignore_index=True)
print (df)
   col1 col2
0     0    a
1     1    D
2     0    b
3     1    E
4     0    c
5     1    F

How it working:

print (df.groupby('col1').cumcount())
0    0
1    1
2    2
3    0
4    1
5    2
dtype: int64