Pandas Set row value based on another column value but do nothing on else-CodePudding

this is my dataframe:

its got 455 rows with a secuence of a period of days in range of 4 hours each row.

i need to replace each 'demand' value with 0 if the timestamp hours are "23"

so i write this:

datadf['value']=datadf['timestamp'].apply(lambda x, y=datadf['value']: 0 if x.hour==23 else y)

i know the Y value is wrong, but i couldnt find the way to refer to the same row "demand" value inside the lambda.

how can i refer to that demand value? is any alternative that my else do nothing?

CodePudding user response：

import pandas as pd
import numpy as np

#data preparation
df = pd.DataFrame()
df['date'] = pd.date_range(start='2022-06-01',periods=7,freq='4h')   pd.Timedelta('3H')
df['val'] = np.random.rand(7)

print(df)
>>
                 date       val
0 2022-06-01 03:00:00  0.601889
1 2022-06-01 07:00:00  0.017787
2 2022-06-01 11:00:00  0.290662
3 2022-06-01 15:00:00  0.179150
4 2022-06-01 19:00:00  0.763534
5 2022-06-01 23:00:00  0.680892
6 2022-06-02 03:00:00  0.585380

#if your dates not datetime format, you must convert it
df['date'] = pd.to_datetime(df['date'])

df.loc[df['date'].dt.hour == 23, 'val'] = 0
#if you don't want to change data in "demand" column you can copy it
#df['val_2'] = df['val']
#df.loc[df['date'].dt.hour == 23, 'val_2'] = 0
print(df)
>>
                 date       val
0 2022-06-01 03:00:00  0.601889
1 2022-06-01 07:00:00  0.017787
2 2022-06-01 11:00:00  0.290662
3 2022-06-01 15:00:00  0.179150
4 2022-06-01 19:00:00  0.763534
5 2022-06-01 23:00:00  0.000000
6 2022-06-02 03:00:00  0.585380