Reputation: 1415
I have a dataframe with a 1 hour periodic DateTime column ts_start
and a shifted by one column ts_end
:
import pandas as pd
from datetime import datetime, timedelta
now = datetime.now()
d = pd.date_range(now, now + timedelta(hours=7), freq='h')
np.random.seed(seed=1111)
df = pd.DataFrame({'ts_start': d, 'col2': range(len(d))})
df = df.assign(ts_end=df['ts_start'].shift(-1))
print(df)
ts_start col2 ts_end
0 2019-05-17 16:51:23.630583 0 2019-05-17 17:51:23.630583
1 2019-05-17 17:51:23.630583 1 2019-05-17 18:51:23.630583
2 2019-05-17 18:51:23.630583 2 2019-05-17 19:51:23.630583
3 2019-05-17 19:51:23.630583 3 2019-05-17 20:51:23.630583
4 2019-05-17 20:51:23.630583 4 2019-05-17 21:51:23.630583
5 2019-05-17 21:51:23.630583 5 2019-05-17 22:51:23.630583
6 2019-05-17 22:51:23.630583 6 2019-05-17 23:51:23.630583
7 2019-05-17 23:51:23.630583 7 NaT
and I'd like to fill NaT with the next hour value, i.e. 2019-05-18 00:51:23.630583
interpolate()
or interpolate(method='time')
do not do anything,
shift(-1, freq='h')
produces:
NotImplementedError: Not supported for type RangeIndex
I am pretty sure there must be something simple to extend a datetime range further.
Upvotes: 1
Views: 1048
Reputation: 317
Try this function:
def fill_in_nat(row):
if pd.isnull(row['ts_end']) == True:
row['ts_end'] = row['ts_start']+timedelta(hours=1)
else:
pass
return row
And then apply it to the DataFrame:
df = df.apply(lambda x: fill_in_nat(x), axis=1)
Upvotes: 1
Reputation: 294258
Add an offset to the shifted column
df.ts_end.fillna(df.ts_end.shift() + pd.offsets.Hour(1))
0 2019-05-17 08:10:39.380197
1 2019-05-17 09:10:39.380197
2 2019-05-17 10:10:39.380197
3 2019-05-17 11:10:39.380197
4 2019-05-17 12:10:39.380197
5 2019-05-17 13:10:39.380197
6 2019-05-17 14:10:39.380197
7 2019-05-17 15:10:39.380197
Name: ts_end, dtype: datetime64[ns]
Upvotes: 2