只需访问Series.dt.isocalendar()的week属性：
示例：

In [286]:
df['Date'].dt.isocalendar().week

Out[286]:
0    25
dtype: int64

In [287]:
df['Week_Number'] = df['Date'].dt.isocalendar().week
df

Out[287]:
        Date  Week_Number
0 2015-06-17           25

这里是使用strftime的另一种可能性。strftime.org是一个很好的资源。

df['Week_Number'] = df['Date'].dt.strftime('%U')

'%U'表示一年中的周数（星期日为该周的第一天），以零填充的十进制数字表示。新的一年中第一个星期日之前的所有天都被视为第0周。
如果您有多个年份的日期，我建议您创建年-周组合

df['Year-Week'] = df['Date'].dt.strftime('%Y-%U')

Pandas具有.dayofyear和.weekofyear功能，可直接应用于pandas.to_datetime(df['column_name'])的输出，并给出类型“时间戳”作为输出。

import pandas as pd
df['formatted_date'] = pd.to_datetime(df['datetime'])
df['day_of_year'] = df.formatted_date.apply(lambda x: x.dayofyear)
df['week_of_year'] = df.formatted_date.apply(lambda x: x.weekofyear)

from datetime import date
df_date = pd.DataFrame([date.today()],columns  = ['today'])
print(df_date)
#### Print Output ####
#        today
#0  2019-09-07
df_date['weeknum'] = df_date.today.apply(lambda x:x.isocalendar()[1])
print(df_date)
#### Print Output ####
#        today  weeknum
#0  2019-09-07       36

更新此答案

在我当前的python版本（3.7，2021年5月）中，语法df['Date'].dt.week打印了以下警告：FutureWarning: weekofyear and week have been deprecated, please use DatetimeIndex.isocalendar().week instead使用日期时间索引的方法是：df['week_number'] = pd.DatetimeIndex(df.index).isocalendar().week
这里有一个小小的演示，用它来返回一个Series

# Input
time_idx = pd.date_range('2022-01-01', periods=4, freq='H').tz_localize('UTC')
values = [9 , 8, 7, 6]

df1 = pd.DataFrame(data = values, index=time_idx, columns=['vals'])
# FutureWarning: weekofyear and week have been deprecated
df1['week_number'] = df1.index.week 

# Using DatetimeIndex.isocalendar().week instead
df2 = pd.DataFrame(data = values, index=time_idx, columns=['vals']) 
# Does not throws a warning
df2['week_number'] = pd.DatetimeIndex(df2.index).isocalendar().week 

print(df2)

如果是Pandas：

import random
import pandas as pd

desired_length = 100
desired_frequency="20D" # XXXM: XXX months, "XXXD":XXX days, XXXMin: XXX minutes etc.

index = pd.date_range('2020-01-01', periods=desired_length, freq=desired_frequency)
data = [random.random() for _ in range(len(index))]

df = pd.DataFrame(data=data, index=index, columns=['DATA'])
df[df.index.isocalendar().keys()] = df.index.isocalendar()