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Python time week-number

该月的第几周？

发布于 2024-09-25 06:43:25 字数 41 浏览 3 评论 0原文

python 是否提供了一种轻松获取当月当前周 (1:4) 的方法？

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烏雲後面有陽光 2024-10-02 06:43:25

为了使用直除法，您要查看的日期所在的月份需要根据该月第一天的位置（在一周内）进行调整。因此，如果您的月份恰好从星期一（一周的第一天）开始，您可以按照上面的建议进行除法。但是，如果该月从星期三开始，您需要加 2，然后进行除法。这一切都封装在下面的函数中。

from math import ceil

def week_of_month(dt):
    """ Returns the week of the month for the specified date.
    """

    first_day = dt.replace(day=1)

    dom = dt.day
    adjusted_dom = dom + first_day.weekday()

    return int(ceil(adjusted_dom/7.0))

In order to use straight division, the day of month for the date you're looking at needs to be adjusted according to the position (within the week) of the first day of the month. So, if your month happens to start on a Monday (the first day of the week), you can just do division as suggested above. However, if the month starts on a Wednesday, you'll want to add 2 and then do the division. This is all encapsulated in the function below.

from math import ceil

def week_of_month(dt):
    """ Returns the week of the month for the specified date.
    """

    first_day = dt.replace(day=1)

    dom = dt.day
    adjusted_dom = dom + first_day.weekday()

    return int(ceil(adjusted_dom/7.0))

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她如夕阳 2024-10-02 06:43:25

我知道这已经有很多年了，但我花了很多时间试图找到这个答案。我制定了自己的方法并认为我应该分享。

日历模块有一个monthcalendar方法，它返回一个二维数组，其中每行代表一周。例如：

import calendar
calendar.monthcalendar(2015,9)

结果：

[[0,0,1,2,3,4,5],
 [6,7,8,9,10,11,12],
 [13,14,15,16,17,18,19],
 [20,21,22,23,24,25,26],
 [27,28,29,30,0,0,0]]

那么 numpy 的结果是你的朋友在哪里。我在美国，所以我希望这一周从周日开始，并将第一周标记为 1：

import calendar
import numpy as np
calendar.setfirstweekday(6)

def get_week_of_month(year, month, day):
    x = np.array(calendar.monthcalendar(year, month))
    week_of_month = np.where(x==day)[0][0] + 1
    return(week_of_month)

get_week_of_month(2015,9,14)

返回

I know this is years old, but I spent a lot of time trying to find this answer. I made my own method and thought I should share.

The calendar module has a monthcalendar method that returns a 2D array where each row represents a week. For example:

import calendar
calendar.monthcalendar(2015,9)

result:

[[0,0,1,2,3,4,5],
 [6,7,8,9,10,11,12],
 [13,14,15,16,17,18,19],
 [20,21,22,23,24,25,26],
 [27,28,29,30,0,0,0]]

So numpy's where is your friend here. And I'm in USA so I want the week to start on Sunday and the first week to be labelled 1:

import calendar
import numpy as np
calendar.setfirstweekday(6)

def get_week_of_month(year, month, day):
    x = np.array(calendar.monthcalendar(year, month))
    week_of_month = np.where(x==day)[0][0] + 1
    return(week_of_month)

get_week_of_month(2015,9,14)

returns

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酒浓于脸红 2024-10-02 06:43:25

如果您的第一周从该月的第一天开始，您可以使用整数除法：

import datetime
day_of_month = datetime.datetime.now().day
week_number = (day_of_month - 1) // 7 + 1

If your first week starts on the first day of the month you can use integer division:

import datetime
day_of_month = datetime.datetime.now().day
week_number = (day_of_month - 1) // 7 + 1

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﹎☆浅夏丿初晴 2024-10-02 06:43:25

查看包 Pendulum

>>> dt = pendulum.parse('2018-09-30')
>>> dt.week_of_month
5

Check out the package Pendulum

>>> dt = pendulum.parse('2018-09-30')
>>> dt.week_of_month
5

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暖心男生 2024-10-02 06:43:25

这个版本可以改进，但作为 python 模块（日期时间和日历）的第一眼，我提出了这个解决方案，我希望有用：

from datetime import datetime
n = datetime.now()
#from django.utils.timezone import now
#n = now() #if you use django with timezone

from calendar import Calendar
cal = Calendar() # week starts Monday
#cal = Calendar(6) # week stars Sunday

weeks = cal.monthdayscalendar(n.year, n.month)
for x in range(len(weeks)):
    if n.day in weeks[x]:
        print x+1

This version could be improved but as a first look in python modules (datetime and calendar), I make this solution, I hope could be useful:

from datetime import datetime
n = datetime.now()
#from django.utils.timezone import now
#n = now() #if you use django with timezone

from calendar import Calendar
cal = Calendar() # week starts Monday
#cal = Calendar(6) # week stars Sunday

weeks = cal.monthdayscalendar(n.year, n.month)
for x in range(len(weeks)):
    if n.day in weeks[x]:
        print x+1

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Saygoodbye 2024-10-02 06:43:25

def week_of_month(date_value):
    week = date_value.isocalendar()[1] - date_value.replace(day=1).isocalendar()[1] + 1
    return date_value.isocalendar()[1] if week < 0 else week

date_value 应采用时间戳格式
这将在所有情况下给出完美的答案。它纯粹基于 ISO 日历

def week_of_month(date_value):
    week = date_value.isocalendar()[1] - date_value.replace(day=1).isocalendar()[1] + 1
    return date_value.isocalendar()[1] if week < 0 else week

date_value should in timestamp format
This will give the perfect answer in all the cases. It is purely based on ISO calendar

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维持三分热 2024-10-02 06:43:25

乔希的答案必须稍作调整，以适应第一天是周日。

def get_week_of_month(date):
   first_day = date.replace(day=1)

   day_of_month = date.day

   if(first_day.weekday() == 6):
       adjusted_dom = (1 + first_day.weekday()) / 7
   else:
       adjusted_dom = day_of_month + first_day.weekday()

   return int(ceil(adjusted_dom/7.0))

Josh's answer has to be tweaked slightly to accomodate the first day falling on a Sunday.

def get_week_of_month(date):
   first_day = date.replace(day=1)

   day_of_month = date.day

   if(first_day.weekday() == 6):
       adjusted_dom = (1 + first_day.weekday()) / 7
   else:
       adjusted_dom = day_of_month + first_day.weekday()

   return int(ceil(adjusted_dom/7.0))

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早乙女 2024-10-02 06:43:25

数据['wk_of_mon'] = (数据['dataset_date'].dt.day - 1) // 7 + 1

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困倦 2024-10-02 06:43:25

查看 python 日历模块

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我的影子我的梦 2024-10-02 06:43:25

我找到了一个非常简单的方法：

import datetime
def week(year, month, day):
    first_week_month = datetime.datetime(year, month, 1).isocalendar()[1]
    if month == 1 and first_week_month > 10:
        first_week_month = 0
    user_date = datetime.datetime(year, month, day).isocalendar()[1]
    if month == 1 and user_date > 10:
        user_date = 0
    return user_date - first_week_month

如果是第一周则返回 0

I found a quite simple way:

import datetime
def week(year, month, day):
    first_week_month = datetime.datetime(year, month, 1).isocalendar()[1]
    if month == 1 and first_week_month > 10:
        first_week_month = 0
    user_date = datetime.datetime(year, month, day).isocalendar()[1]
    if month == 1 and user_date > 10:
        user_date = 0
    return user_date - first_week_month

returns 0 if first week

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骄兵必败 2024-10-02 06:43:25

乔什的答案似乎是最好的，但我认为我们应该考虑这样一个事实：只有当周四属于该月时，一周才属于该月。至少iso 是这么说的。

根据该标准，一个月最多可以有 5 周。一天可以属于一个月，但它所属的一周可能不属于。

我已经考虑到，只需添加一个简单的

if (first_day.weekday()>3) :
        return ret_val-1
    else:
        return ret_val

其中 ret_val 正是 Josh 的计算值。于 2017 年 6 月（有 5 周）和 2017 年 9 月进行了测试。通过“2017-09-01”将返回 0，因为该天属于不属于 9 月的一周。

最正确的方法是让该方法返回输入日期所属的周数和月份名称。

Josh' answer seems the best but I think that we should take into account the fact that a week belongs to a month only if its Thursday falls into that month. At least that's what the iso says.

According to that standard, a month can have up to 5 weeks. A day could belong to a month, but the week it belongs to may not.

I have taken into account that just by adding a simple

if (first_day.weekday()>3) :
        return ret_val-1
    else:
        return ret_val

where ret_val is exactly Josh's calculated value. Tested on June 2017 (has 5 weeks) and on September 2017. Passing '2017-09-01' returns 0 because that day belongs to a week that does not belong to September.

The most correct way would be to have the method return both the week number and the month name the input day belongs to.

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温柔戏命师 2024-10-02 06:43:25

@Manuel Solorzano 的答案的变体：

from calendar import monthcalendar
def get_week_of_month(year, month, day):
    return next(
        (
            week_number
            for week_number, days_of_week in enumerate(monthcalendar(year, month), start=1)
            if day in days_of_week
        ),
        None,
    )

例如：

>>> get_week_of_month(2020, 9, 1)
1
>>> get_week_of_month(2020, 9, 30)
5
>>> get_week_of_month(2020, 5, 35)
None

A variation on @Manuel Solorzano's answer:

from calendar import monthcalendar
def get_week_of_month(year, month, day):
    return next(
        (
            week_number
            for week_number, days_of_week in enumerate(monthcalendar(year, month), start=1)
            if day in days_of_week
        ),
        None,
    )

E.g.:

>>> get_week_of_month(2020, 9, 1)
1
>>> get_week_of_month(2020, 9, 30)
5
>>> get_week_of_month(2020, 5, 35)
None

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热鲨 2024-10-02 06:43:25

假设我们有某个月的日历，如下所示：

Mon Tue Wed Thur Fri Sat Sun
                 1   2   3 
4   5   6   7    8   9   10

我们说第 1 ~ 3 天属于第 1 周，第 4 ~ 10 天属于第 2 周等等。

在这种情况下，我相信特定日期的 week_of_month 应该计算如下：

import datetime
def week_of_month(year, month, day):
    weekday_of_day_one = datetime.date(year, month, 1).weekday()
    weekday_of_day = datetime.date(year, month, day)
    return (day - 1)//7 + 1 + (weekday_of_day < weekday_of_day_one)

但是，如果我们想要获取该日期所在工作日的第 n 个，例如第 1 天是第 1 个星期五，第 8 天是第 2 个星期五星期五，第 6 天是第一个星期三，那么我们可以简单地返回 (day - 1)//7 + 1

Say we have some month's calender as follows:

Mon Tue Wed Thur Fri Sat Sun
                 1   2   3 
4   5   6   7    8   9   10

We say day 1 ~ 3 belongs to week 1 and day 4 ~ 10 belongs to week 2 etc.

In this case, I believe the week_of_month for a specific day should be calculated as follows:

import datetime
def week_of_month(year, month, day):
    weekday_of_day_one = datetime.date(year, month, 1).weekday()
    weekday_of_day = datetime.date(year, month, day)
    return (day - 1)//7 + 1 + (weekday_of_day < weekday_of_day_one)

However, if instead we want to get the nth of the weekday that date is, such as day 1 is the 1st Friday, day 8 is the 2nd Friday, and day 6 is the 1st Wednesday, then we can simply return (day - 1)//7 + 1

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靖瑶 2024-10-02 06:43:25

您正在寻找的答案是(dm-dw+(dw-dm)%7)/7+1，其中dm是该月的日期，dw 是星期几，% 是正余数。

这来自于月份偏移量 (mo) 和月份中的第几周 (wm) 的关联，其中月份偏移量是第一天开始的时间。如果我们考虑所有这些变量都从 0 开始，

wm*7+dw = dm+mo

您可以求解 mo 的模 7，因为这会导致 wm 变量丢失，因为它只显示为倍数7

dw = dm+mo   (%7)
mo = dw-dm   (%7)
mo = (dw-dm)%7  (since the month offset is 0-6)

然后您只需将月份偏移量代入原始方程并求解 wm

wm*7+dw = dm+mo
wm*7 = dm-dw + mo
wm*7 = dm-dw + (dw-dm)%7
wm = (dm-dw + (dw-dm)%7) / 7

由于 dm 和 dw 始终成对出现，因此可以将它们偏移任何数量，因此，将所有内容切换为从 1 开始只会将等式更改为 (dm-dw + (dw-dm)%7)/7 + 1。

当然，python datetime 库从 1 开始 dm ，从 0 开始 dw 。因此，假设 date 是一个 < code>datatime.date 对象，您可以使用

(date.day-1-date.dayofweek() + (date.dayofweek()-date.day+1)%7) / 7 + 1

由于内部位始终是 7 的倍数（字面上是 dw*7），您可以看到第一个 -date.dayofweek() 只是将值向后调整为最接近的 7 倍数。整数除法也可以执行此操作，因此可以进一步简化为

(date.day-1 + (date.dayofweek()-date.day+1)%7) // 7 + 1

注意 dayofweek() 将星期日置于周末。

The answer you are looking for is (dm-dw+(dw-dm)%7)/7+1 where dm is the day of the month, dw is the day of the week, and % is the positive remainder.

This comes from relating the month offset (mo) and the week of the month (wm), where the month offset is how far into the week the first day starts. If we consider all of these variables to start at 0 we have

wm*7+dw = dm+mo

You can solve this modulo 7 for mo as that causes the wm variable drops out as it only appears as a multiple of 7

dw = dm+mo   (%7)
mo = dw-dm   (%7)
mo = (dw-dm)%7  (since the month offset is 0-6)

Then you just substitute the month offset into the original equation and solve for wm

wm*7+dw = dm+mo
wm*7 = dm-dw + mo
wm*7 = dm-dw + (dw-dm)%7
wm = (dm-dw + (dw-dm)%7) / 7

As dm and dw are always paired, these can be offset by any amount, so, switching everything to start a 1 only changes the the equation to (dm-dw + (dw-dm)%7)/7 + 1.

Of course the python datetime library starts dm at 1 and dw at 0. So, assuming date is a datatime.date object, you can go with

(date.day-1-date.dayofweek() + (date.dayofweek()-date.day+1)%7) / 7 + 1

As the inner bit is always a multiple of 7 (it is literally dw*7), you can see that the first -date.dayofweek() simply adjusts the value backwards to closest multiple of 7. Integer division does this too, so it can be further simplified to

(date.day-1 + (date.dayofweek()-date.day+1)%7) // 7 + 1

Be aware that dayofweek() puts Sunday at the end of the week.

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冷血 2024-10-02 06:43:25

这应该可以做到。

#! /usr/bin/env python2

import calendar, datetime

#FUNCTIONS
def week_of_month(date):
    """Determines the week (number) of the month"""

    #Calendar object. 6 = Start on Sunday, 0 = Start on Monday
    cal_object = calendar.Calendar(6)
    month_calendar_dates = cal_object.itermonthdates(date.year,date.month)

    day_of_week = 1
    week_number = 1

    for day in month_calendar_dates:
        #add a week and reset day of week
        if day_of_week > 7:
            week_number += 1
            day_of_week = 1

        if date == day:
            break
        else:
            day_of_week += 1

    return week_number


#MAIN
example_date = datetime.date(2015,9,21)

print "Week",str(week_of_month(example_date))
#Returns 'Week 4'

This should do it.

#! /usr/bin/env python2

import calendar, datetime

#FUNCTIONS
def week_of_month(date):
    """Determines the week (number) of the month"""

    #Calendar object. 6 = Start on Sunday, 0 = Start on Monday
    cal_object = calendar.Calendar(6)
    month_calendar_dates = cal_object.itermonthdates(date.year,date.month)

    day_of_week = 1
    week_number = 1

    for day in month_calendar_dates:
        #add a week and reset day of week
        if day_of_week > 7:
            week_number += 1
            day_of_week = 1

        if date == day:
            break
        else:
            day_of_week += 1

    return week_number


#MAIN
example_date = datetime.date(2015,9,21)

print "Week",str(week_of_month(example_date))
#Returns 'Week 4'

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假面具 2024-10-02 06:43:25

移至该月一周的最后一天并除以 7

from math import ceil

def week_of_month(dt):
    """ Returns the week of the month for the specified date.
    """
    # weekday from monday == 0 ---> sunday == 6
    last_day_of_week_of_month = dt.day + (7 - (1 + dt.weekday()))
    return int(ceil(last_day_of_week_of_month/7.0))

Move to last day of week in month and divide to 7

from math import ceil

def week_of_month(dt):
    """ Returns the week of the month for the specified date.
    """
    # weekday from monday == 0 ---> sunday == 6
    last_day_of_week_of_month = dt.day + (7 - (1 + dt.weekday()))
    return int(ceil(last_day_of_week_of_month/7.0))

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岁月静好 2024-10-02 06:43:25

您可以简单地执行以下操作：

首先提取月份和年份中的周数

df['month'] = df['Date'].dt.month
df['week'] = df['Date'].dt.week

然后按月分组并对周数进行排名

df['weekOfMonth'] = df.groupby('month')["week"].rank("dense", ascending=False)

You can simply do as follow:

First extract the month and the week of year number

df['month'] = df['Date'].dt.month
df['week'] = df['Date'].dt.week

Then group by month and rank the week numbers

df['weekOfMonth'] = df.groupby('month')["week"].rank("dense", ascending=False)

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静谧幽蓝 2024-10-02 06:43:25

还有一个解决方案，其中星期日是一周的第一天，仅基于 Python。

def week_of_month(dt):
""" Returns the week of the month for the specified date.
TREATS SUNDAY AS FIRST DAY OF WEEK!
"""
    first_day = dt.replace(day=1)
    dom = dt.day
    adjusted_dom = dom + (first_day.weekday() + 1) % 7
    return (adjusted_dom - 1) // 7 + 1

One more solution, where Sunday is first day of week, base Python only.

def week_of_month(dt):
""" Returns the week of the month for the specified date.
TREATS SUNDAY AS FIRST DAY OF WEEK!
"""
    first_day = dt.replace(day=1)
    dom = dt.day
    adjusted_dom = dom + (first_day.weekday() + 1) % 7
    return (adjusted_dom - 1) // 7 + 1

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丢了幸福的猪 2024-10-02 06:43:25

def week_number(time_ctime = None):
    import time
    import calendar
    if time_ctime == None:
        time_ctime = str(time.ctime())
    date = time_ctime.replace('  ',' ').split(' ')
    months = {'Jan':1,'Feb':2,'Mar':3,'Apr':4,'May':5,'Jun':6,'Jul':7,'Aug':8,'Sep':9,'Oct':10,'Nov':11,'Dec':12}
    week, day, month, year = (-1, str(date[2]), months[date[1]], int(date[-1]))
    cal = calendar.monthcalendar(year,month)
    for wk in range(len(cal)):
        wstr = [str(x) for x in cal[wk]]
        if day in wstr:
            week = wk
            break
    return week

import time
print(week_number())
print(week_number(time.ctime()))

def week_number(time_ctime = None):
    import time
    import calendar
    if time_ctime == None:
        time_ctime = str(time.ctime())
    date = time_ctime.replace('  ',' ').split(' ')
    months = {'Jan':1,'Feb':2,'Mar':3,'Apr':4,'May':5,'Jun':6,'Jul':7,'Aug':8,'Sep':9,'Oct':10,'Nov':11,'Dec':12}
    week, day, month, year = (-1, str(date[2]), months[date[1]], int(date[-1]))
    cal = calendar.monthcalendar(year,month)
    for wk in range(len(cal)):
        wstr = [str(x) for x in cal[wk]]
        if day in wstr:
            week = wk
            break
    return week

import time
print(week_number())
print(week_number(time.ctime()))

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花心好男孩 2024-10-02 06:43:25

获取月份周数的简单方法；

如果数据类型是 datetime64 那么

week_number_of_month = date_value.dayofweek

An Easy way to get a week number of month;

if the datatype is datetime64 then

week_number_of_month = date_value.dayofweek

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谁对谁错谁最难过 2024-10-02 06:43:25

我发现的最好的解决方案是这个函数

def get_week_of_month(date):
date = datetime.strptime(date, "%Y-%m-%d") # date is str
first_day = date.replace(day=1)
print(first_day)
day_of_month = date.day
print(day_of_month)
print(first_day.weekday())
if(first_day.weekday() == 6):
    adjusted_dom = day_of_month + ((1 + first_day.weekday()) / 7)
else:
    adjusted_dom = day_of_month + first_day.weekday()
print(adjusted_dom)
return int(ceil(adjusted_dom/7.0))

the best solution that I found is this function

def get_week_of_month(date):
date = datetime.strptime(date, "%Y-%m-%d") # date is str
first_day = date.replace(day=1)
print(first_day)
day_of_month = date.day
print(day_of_month)
print(first_day.weekday())
if(first_day.weekday() == 6):
    adjusted_dom = day_of_month + ((1 + first_day.weekday()) / 7)
else:
    adjusted_dom = day_of_month + first_day.weekday()
print(adjusted_dom)
return int(ceil(adjusted_dom/7.0))

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红颜悴 2024-10-02 06:43:25

  import datetime
    
  def week_number_of_month(date_value):
            week_number = (date_value.isocalendar()[1] - date_value.replace(day=1).isocalendar()[1] + 1)
            if week_number == -46:
                week_number = 6
           return week_number
                
 date_given = datetime.datetime(year=2018, month=12, day=31).date()
                
 week_number_of_month(date_given)

  import datetime
    
  def week_number_of_month(date_value):
            week_number = (date_value.isocalendar()[1] - date_value.replace(day=1).isocalendar()[1] + 1)
            if week_number == -46:
                week_number = 6
           return week_number
                
 date_given = datetime.datetime(year=2018, month=12, day=31).date()
                
 week_number_of_month(date_given)

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霓裳挽歌倾城醉 2024-10-02 06:43:25

此解决方案旨在复制 Python 中 WeekOfMonth 的 Java 实现，并遵循类似于 ISO 8601 约定。

一周的第一天是星期一

第一周的最少天数是 4。这意味着从星期一到星期四（含）之间开始的周是整周

 来自日期时间导入日期

    def week_of_month(calendar_date: 日期) ->整数：
    ”“”
      Python 实现月份中的一周
      遵循 ISO 8601 https://en.wikipedia.org/wiki/ISO_week_date，

      1.一周的第一天是星期一
      2. 第一周最少天数为 4 天
          从周一到周四开始的周
          包括整周

      该函数返回周数
      日历日期一个月内

      ：参数日历日期：
      :return: 一个月中的第几周
    ”“”
    _、月、_ = 日历日期.年、\
                  日历_日期.月份，\
                  日历日期.day
    _, week_of_year, _ = calendar_date.isocalendar()

    date_at_month_start = calendar_date.replace(day=1)
    _，月份_开始_周_年，月份_开始_日_周 = \
        date_at_month_start.isocalendar()

    如果月份 == 1：
        如果 week_of_year >= 51，则返回 0，否则 week_of_year

    elif Month_start_day_of_week > > 4：
        返回year_year - Month_start_week_of_year

    别的：
        返回year_year - Month_start_week_of_year + 1

This solution was intended to replicate the Java implementation of WeekOfMonth in Python and follow a pattern similar to the ISO 8601 convention.

First day of the week is Monday

Minimal number of days in the first week is 4. This implies weeks starting on and between Monday and Thursday inclusive are full weeks

    from datetime import date

    def week_of_month(calendar_date: date) -> int:
    """
      Python implementation of Week of Month
      following ISO 8601 https://en.wikipedia.org/wiki/ISO_week_date,

      1. First day of the week is Monday
      2. Minimal days in the first week is 4 so
          weeks starting on and between Monday and Thursday
          inclusive are full weeks

      This function returns the week's number
      within a month for a calendar date

      :param calendar_date:
      :return: week of the month
    """
    _, month, _ = calendar_date.year, \
                  calendar_date.month, \
                  calendar_date.day
    _, week_of_year, _ = calendar_date.isocalendar()

    date_at_month_start = calendar_date.replace(day=1)
    _, month_start_week_of_year, month_start_day_of_week = \
        date_at_month_start.isocalendar()

    if month == 1:
        return 0 if week_of_year >= 51 else week_of_year

    elif month_start_day_of_week > 4:
        return week_of_year - month_start_week_of_year

    else:
        return week_of_year - month_start_week_of_year + 1

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