Pandas: preencha nan usando o valor anterior e interpolando

Acho que preciso preencher com interpolar:
df ['colB'] = df ['col_A']. ffill ()
df ['colc'] = df ['col_A']. interpolar ()
imprimir (df)
hora col_A colB colc
0 1.520583e + 09 79.000 79.000 79.00000
1 1.520583e + 09 22.500 22.500 22.50000
2 1.520583e + 09 29.361 29.361 29.36100
3 1.520583e + 09 116.095 116.095 116.09500
4 1.520584e + 09 19.972 19.972 19,97200
5 1.520584e + 09 36.857 36.857 36.85700
6 1.520584e + 09 15.167 15.167 15.16700
7 1.520585e + 09 NaN 15.167 20.00025
8 1,520585e + 09 NaN 15,167 24,83350
9 1,520585e + 09 NaN 15,167 29,66675
10 1.520586e + 09 34.500 34.500 34.50000
11 1.520586e + 09 17.583 17.583 17.58300
12 1.520586e + 09 NaN 17.583 33.20800
13 1.520586e + 09 48.833 48.833 48.83300
14 1.520587e + 09 18.806 18.806 18.80600
15 1.520587e + 09 18.583 18.583 18.58300
Se quiser usar o método tempo para interpolar:
df ['hora'] = pd.to_datetime (df ['hora'], unidade = 's')
df = df.set_index ('tempo')
df ['colB'] = df ['col_A']. ffill ()
df ['colc'] = df ['col_A']. interpolar ('tempo')
imprimir (df)
col_A colB colc
Tempo
09-03-2018 08:03:00 79.000 79.000 79.00000
2018-03-09 08:08:00 22.500 22.500 22.50000
2018-03-09 08:13:00 29.361 29.361 29.36100
2018-03-09 08:18:00 116.095 116.095 116.09500
2018-03-09 08:23:00 19.972 19.972 19,97200
2018-03-09 08:28:00 36.857 36.857 36.85700
2018-03-09 08:33:00 15.167 15.167 15.16700
09-03-2018 08:38:00 NaN 15.167 20.00025
2018-03-09 08:43:00 NaN 15.167 24.83350
2018-03-09 08:48:00 NaN 15.167 29.66675
2018-03-09 08:53:00 34.500 34.500 34.50000
2018-03-09 08:58:00 17.583 17.583 17.58300
2018-03-09 09:03:00 NaN 17.583 33.20800
2018-03-09 09:08:00 48.833 48.833 48.83300
2018-03-09 09:13:00 18.806 18.806 18.80600
2018-03-09 09:18:00 18.583 18.583 18.58300
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