使用python获取两个日期之间的周数

类似于( 更新: 删除了可读性较差的选项)

import datetime

def find_weeks(start,end):
    l = []
    for i in range((end-start).days + 1):
        d = (start+datetime.timedelta(days=i)).isocalendar()[:2] # e.g. (2011, 52)
        yearweek = '{}{:02}'.format(*d) # e.g. "201152"
        l.append(yearweek)
    return sorted(set(l))

start = datetime.date(2011, 12, 25) 
end = datetime.date(2012, 1, 21)

print(find_weeks(start,end)[1:]) # [1:] to exclude first week.

退换商品

['201152', '201201', '201202', '201203']

要包括第一周(201151),只需删除 [1:] 函数调用后

.isocalendar() 这里是您的朋友-它返回一个元组 (year, week of year, day of week) . 我们使用它将开始日期重置为eweek的开始日期,然后每次增加一周,直到我们通过结束日期:

import datetime


def find_weeks(start_date, end_date):
    subtract_days = start_date.isocalendar()[2] - 1
    current_date = start_date + datetime.timedelta(days=7-subtract_days)
    weeks_between = []
    while current_date <= end_date:
        weeks_between.append(
            '{}{:02d}'.format(*current_date.isocalendar()[:2])
        )
        current_date += datetime.timedelta(days=7)
    return weeks_between

start = datetime.date(2011, 12, 25)
end = datetime.date(2012, 1, 21)

print(find_weeks(start, end))

这张照片

['201152', '201201', '201202', '201203']

使用熊猫

import pandas as pd

dates=pd.date_range(start=start,end=end,freq='W')
date_index=dates.year.astype(str)+dates.weekofyear.astype(str).str.zfill(2)
date_index.tolist()

我建议您使用以下易于阅读的解决方案:

import datetime

start = datetime.date(2011, 12, 25) 
end = datetime.date(2012, 1, 21)

def find_weeks(start, end):
    l = []
    while (start.isocalendar()[1] != end.isocalendar()[1]) or (start.year != end.year):
        l.append(start.isocalendar()[1] + 100*start.year)
        start += datetime.timedelta(7)
    l.append(start.isocalendar()[1] + 100*start.year)
    return (l[1:])


print(find_weeks(start, end))

>> [201252, 201201, 201202, 201203]