方法1：转换为聚合和广播值

g = df.groupby('UserID')
df['min_login_date'] = g['LoginDate'].transform('min')
df['max_order_items'] = g['PlacedOrderItems'].transform('max')

方法2：进行命名聚合，然后合并回来

d = {'min_login_date': ('LoginDate', 'min'), 'max_order_items': ('PlacedOrderItems', 'max')}
df = df.merge(df.groupby('UserID').agg(**d))

UserID   LoginDate  PlacedOrderItems Other column min_login_date  max_order_items
0  A123456  2023-04-01                 1            a     2023-04-01                1
1  A123456  2023-04-10                 0            b     2023-04-01                1
2  B123456  2023-04-05                 0            c     2023-04-05                2
3  B123456  2023-04-06                 2            d     2023-04-05                2
4  B123456  2023-04-07                 0            e     2023-04-05                2

你可以通过使用groupby（）方法和transform（）方法创建两个新列来实现这一点：min_login_date和max_order_items：

import pandas as pd

# set display options
pd.set_option('display.max_columns', 10)
pd.set_option('display.width', 100)

# create the dataframe
data = {'UserID': ['A123456', 'A123456', 'B123456', 'B123456', 'B123456'],
        'LoginDate': ['2023-04-01', '2023-04-10', '2023-04-05', '2023-04-06', '2023-04-07'],
        'PlacedOrderItems': [1, 0, 0, 2, 0],
        'OtherColumn': ['a', 'b', 'c', 'd', 'e']}
df = pd.DataFrame(data)

# convert LoginDate to datetime format
df['LoginDate'] = pd.to_datetime(df['LoginDate'])

# group by UserID and transform to get the min LoginDate and max PlacedOrderItems
df['min_login_date'] = df.groupby('UserID')['LoginDate'].transform('min')
df['max_order_items'] = df.groupby('UserID')['PlacedOrderItems'].transform('max')

# group by all columns and take the first row of each group
group_cols = ['UserID', 'LoginDate', 'PlacedOrderItems', 'OtherColumn', 'min_login_date', 'max_order_items']
df = df.groupby(group_cols).first().reset_index()

# print the resulting dataframe
print(df.to_string(index=False))