根据日期列获取周末答案

【问题标题】：Get end of week based on date column根据日期列获取周末
【发布时间】：2015-09-15 04:03:08
【问题描述】：

在下面的数据框中，“end_of_week”列尚不存在。我试图使如果“日期”小于或等于该周的星期四，则“end_of_week”是星期四。我该怎么做呢？

我正在尝试做的事情：

df['end_of_week'] = Thursday of same week if df['date'] <= Thursday

例子：

0   date    end_of_week
1   2015-08-31  2015-09-03 #if <= Thursday of that week
2   2015-09-01  2015-09-03
3   2015-09-07  2015-09-10
4   2015-09-09  2015-09-10
5   2015-09-16  2015-09-17
6   2015-09-17  2015-09-17

谢谢。

【问题讨论】：

标签： python pandas

【解决方案1】：

怎么样：

df['end_of_week'] = df['date'].map(lambda x: next_weekday(x, 3) if  x.weekday() < 3 else x)

哪里，来自@phihag 在Find the date for the first Monday after a given a date 的回答：

def next_weekday(d, weekday):
    days_ahead = weekday - d.weekday()
    if days_ahead <= 0:
        days_ahead += 7
    return d + datetime.timedelta(days_ahead)

做一个测试：

import pandas as pd
import datetime
xl3 = pd.ExcelFile('test2.xlsx')
df3 = xl3.parse("Sheet1")
df3
Out[71]: 
   x          y       date 
0  1        fum 2015-06-01  
1  2         fo 2015-06-02 
2  3         fi 2015-06-03 
3  4        fee 2015-06-04 
4  5  dumbledum 2015-06-05 
5  6  dumbledee 2015-06-06 

df3['end_of_week'] = df3['date'].map(lambda x: next_weekday(x, 3) if x.weekday() < 3 else x)

df3
   x          y       date end_of_week 
0  1        fum 2015-06-01  2015-06-04
1  2         fo 2015-06-02  2015-06-04
2  3         fi 2015-06-03  2015-06-04
3  4        fee 2015-06-04  2015-06-04
4  5  dumbledum 2015-06-05  2015-06-05
5  6  dumbledee 2015-06-06  2015-06-06

【讨论】：

谢谢！！这完全有效。你回答的就像我问的所有问题一样！

【解决方案2】：

您可以使用arrow 查找最近的星期四。 Arrow 认为星期一是一周的第 0 天。

import arrow

THURSDAY = 3
arw = arrow.get("2015-08-31")
arw = arw.replace(days= +(THURSDAY - arw.weekday())).format('YYYY-MM-DD')

#=> arw = '2015-09-03'

直接解决您的问题：

import arrow

THURSDAY = 3
dates = [
    "2015-08-31",
    "2015-09-01",
    "2015-09-07",
    "2015-09-09",
    "2015-09-16",
    "2015-09-17",
]

end_of_week_dates = []

for date in dates:
    arw = arrow.get(date)
    if arw.weekday() <= THURSDAY:
        end_of_week = arw.replace(days= +(THURSDAY - arw.weekday())).format('YYYY-MM-DD')
        end_of_week_dates.append(end_of_week)
    else:
        # handle the case where the date is after the Thursday of that week

# end_of_week_dates = [
#   '2015-09-03',
#   '2015-09-03',
#   '2015-09-10',
#   '2015-09-10',
#   '2015-09-17',
#   '2015-09-17'
# ]

【讨论】：

感谢您的回答！我去了上面Tris的答案