Incrementing datetime object by with a range-CodePudding

I've got the following column in my pandas dataframe. My intention is to increment each hour by 1 within a certain range(0,10). I could easily say: df['time'] = np.arange(0,10) to get those values, however I lose the datetime formatting.

To create the the datetime, I used the following code:

df['time'] = (
    datetime(
        start.year, start.month, start.day, hour=1, tzinfo=timezone.utc
    ))

I am not opposed to a method to just iterate the dataframe and increment them by 1, or create it properly given the range. Any advice would be appreciated.

CodePudding user response：

Possibly not the most elegant, but this sufficed:

for i in range(0, df.shape[0]):
    for j in range(0, 10):
        df['time'][(i*10) j] = df['time'][(i*10) j].replace(hour=j)

CodePudding user response：

You can use timedelta arithmetic. Or use a date_range. Ex:

import pandas as pd
import numpy as np

df = pd.DataFrame()
num = 10
start = pd.Timestamp("now").floor("d") # current date, h/m/s set zero

# add range as timedelta to start date:
df["Time"] = start   pd.to_timedelta(np.arange(num), unit="H")
# even simpler: date_range
df["Time2"] = pd.date_range(start, periods=num, freq="H")


df
                 Time               Time2
0 2022-10-06 00:00:00 2022-10-06 00:00:00
1 2022-10-06 01:00:00 2022-10-06 01:00:00
2 2022-10-06 02:00:00 2022-10-06 02:00:00
3 2022-10-06 03:00:00 2022-10-06 03:00:00
4 2022-10-06 04:00:00 2022-10-06 04:00:00
5 2022-10-06 05:00:00 2022-10-06 05:00:00
6 2022-10-06 06:00:00 2022-10-06 06:00:00
7 2022-10-06 07:00:00 2022-10-06 07:00:00
8 2022-10-06 08:00:00 2022-10-06 08:00:00
9 2022-10-06 09:00:00 2022-10-06 09:00:00