如何在 lambda 中使用 isin

Question

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

在所有“開放時間”值相交的行上尋找匹配項。 您需要將索引值寫入“LEVEL”列

使用代碼： df['LEVEL']=df['intersect'].loc[df['intersect'].isna()==0].apply(lambda x: df[df['Open Time'].isin(x.replace(r'intersect_', ''))].index)

返回錯誤：只允許將類似列表的對象傳遞給 isin()，你傳遞了一個 [str]。 什么是可能的替代品 .isin ？

請告訴我如何使用 lambda 函數編寫類似的代碼： df[df['Open Time'].isin(df['intersect'].str.replace(r'intersect_', ''))].index.tolist()

需要結果：

         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

Answer 1

鑒於：

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})

正在做：

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)

輸出：

                   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]