原问题:单系列值
您可以根据您的条件定义一个布尔系列,然后通过numpy.where 酌情定义interpolate 或ffill:
# setup
df = pd.DataFrame({'date': ['02/03/2016 05:00', '02/03/2016 06:00', '02/03/2016 07:00',
'02/03/2016 08:00', '02/03/2016 09:00'],
'value': [8, np.nan, 1, np.nan, 3]})
df['date'] = pd.to_datetime(df['date'])
# construct Boolean switch series
switch = (df['date'] - df['date'].dt.normalize()) > pd.to_timedelta('07:00:00')
# use numpy.where to differentiate between two scenarios
df['value'] = np.where(switch, df['value'].interpolate(), df['value'].ffill())
print(df)
date value
0 2016-02-03 05:00:00 8.0
1 2016-02-03 06:00:00 8.0
2 2016-02-03 07:00:00 1.0
3 2016-02-03 08:00:00 2.0
4 2016-02-03 09:00:00 3.0
更新问题:多个系列值
对于多个值列,您可以使用pd.DataFrame.where 和iloc 调整上述解决方案。或者,代替iloc,您可以使用loc 或其他方式(例如filter)选择列:
# setup
df = pd.DataFrame({'date': ['02/03/2016 05:00', '02/03/2016 06:00', '02/03/2016 07:00',
'02/03/2016 08:00', '02/03/2016 09:00'],
'value': [8, np.nan, 1, np.nan, 3],
'value2': [3, np.nan, 2, np.nan, 6]})
df['date'] = pd.to_datetime(df['date'])
# construct Boolean switch series
switch = (df['date'] - df['date'].dt.normalize()) > pd.to_timedelta('07:00:00')
# use numpy.where to differentiate between two scenarios
df.iloc[:, 1:] = df.iloc[:, 1:].interpolate().where(switch, df.iloc[:, 1:].ffill())
print(df)
date value value2
0 2016-02-03 05:00:00 8.0 3.0
1 2016-02-03 06:00:00 8.0 3.0
2 2016-02-03 07:00:00 1.0 2.0
3 2016-02-03 08:00:00 2.0 4.0
4 2016-02-03 09:00:00 3.0 6.0