Я буду рекомендовать
df1['addedList']=[['aa', 'bb', 'cc'] for x in range(len(df1))]
df1
Out[156]:
a b addedList
0 a d [aa, bb, cc]
1 b e [aa, bb, cc]
2 c f [aa, bb, cc]
Вот причина, по которой: Если мы создаем столбцы по одной копии, при изменении одного значения все остальные будут изменены.
df1['addedList']=[testList]*df1.shape[0]
df1['addedList1']=[['aa', 'bb', 'cc'] for x in range(len(df1))]
df1.addedList[0][0]=11111
df1
Out[162]:
a b addedList addedList1
0 a d [11111, bb, cc, dd] [aa, bb, cc]
1 b e [11111, bb, cc, dd] [aa, bb, cc]
2 c f [11111, bb, cc, dd] [aa, bb, cc]
df1.addedList1[0][0]=11111
df1
Out[164]:
a b addedList addedList1
0 a d [11111, bb, cc, dd] [11111, bb, cc]
1 b e [11111, bb, cc, dd] [aa, bb, cc]
2 c f [11111, bb, cc, dd] [aa, bb, cc]