import pandas as pd
import numpy as np
Open_Time_s=['2022-04-30 11:05:00 03:00','2022-04-30 11:10:00 03:00', np.nan, np.nan]
intersect=[np.nan,np.nan,'intersect_2022-04-30 11:05:00 03:00','intersect_2022-04-30 11:10:00 03:00']
df = pd.DataFrame.from_dict({'Open Time':Open_Time_s,'intersect':intersect})
df['LEVEL']=np.nan
looking for matches on all lines 'Open Time' values intersect. You need to write the index values to the column 'LEVEL'
Using code : df['LEVEL']=df['intersect'].loc[df['intersect'].isna()==0].apply(lambda x: df[df['Open Time'].isin(x.replace(r'intersect_', ''))].index)
return error : only list-like objects are allowed to be passed to isin(), you passed a [str]. What is the possible alternative .isin ?
please tell me how to write similar code using lambda function:df[df['Open Time'].isin(df['intersect'].str.replace(r'intersect_', ''))].index.tolist()
need result :
Open Time intersect LEVEL
0 2022-04-30 11:05:00 03:00 NaN NaN
1 2022-04-30 11:10:00 03:00 NaN NaN
2 None intersect_2022-04-30 11:05:00 03:00 0
3 None intersect_2022-04-30 11:10:00 03:00 1
CodePudding user response:
Given:
Open_Time_s=['2022-04-30 11:05:00 03:00','2022-04-30 11:10:00 03:00', np.nan, np.nan]
intersect=[np.nan,np.nan,'intersect_2022-04-30 11:05:00 03:00','intersect_2022-04-30 11:10:00 03:00']
df = pd.DataFrame.from_dict({'Open Time':Open_Time_s,'intersect':intersect})
Doing:
df[['intersect', 'intersect_time']] = df['intersect'].str.split('_', expand=True)
df['LEVEL'] = df.apply(lambda x: df[df['Open Time'].eq(x['intersect_time'])].index.tolist(), axis=1)
Output:
Open Time intersect intersect_time LEVEL
0 2022-04-30 11:05:00 03:00 NaN NaN []
1 2022-04-30 11:10:00 03:00 NaN NaN []
2 NaN intersect 2022-04-30 11:05:00 03:00 [0]
3 NaN intersect 2022-04-30 11:10:00 03:00 [1]