python - Week number of the month

问题

Does python offer a way to easily get the current week of the month (1:4) ?

回答1:

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))

回答2:

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

回答3:

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

回答4:

Check out the package Pendulum

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

回答5:

Check out the python calendar module

回答6:

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 now.day in weeks[x]:
        print x+1

回答7:

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))

回答8:

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

回答9:

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.

回答10:

please link to Python: Number of the Week in a Month

IT all depends on when the 1st monday of the month is

回答11:

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'

回答12:

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))

来源：https://stackoverflow.com/questions/3806473/python-week-number-of-the-month

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python

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week-number