CODEWITHSUNDEEP
pythondatetime
Deep Dwivedi
Deep Dwivedi

Reputation: 115

How to find number of Mondays or any other weekday between two dates in Python?

I have two dates between which I need to find out how many Mon- Fri are coming(except for Sta, Sun), everyday should be counted

Currently I am thinking this:

import calendar
import datetime
start_date = datetime.datetime.strptime("01/01/2017",'%d/%m/%Y')
end_date = datetime.datetime.strptime("31/01/2017",'%d/%m/%Y')
week_arr = [0] * 7
calendar.day_name[start_date.weekday()] ## will give me name of day
"""
As I receive Monday I will increment week_arr[0] by 1, Tuesday
week_arr[1]+= 1,
"""

I am not getting how to do it effectively so that I dont use much line of code(less if -else and for loops), may be some tricks in pandas.

Upvotes: 10

Views: 19594

Answers (6)

mohammad
mohammad

Reputation: 2198

You can define a function and use it like this : 

def num_days_between( start, end, week_day):
    num_weeks, remainder = divmod( (end-start).days, 7)
    if ( week_day - start.weekday() ) % 7 < remainder:
       return num_weeks + 1
    else:
       return num_weeks

where week_day is day number you wan to calculate count.

Upvotes: 9

imran Wadkar
imran Wadkar

Reputation: 1

I found a simple and easy to understand code using for loop. Take first date identify its weekday with "%a" compare it with your intrested weekday if found increment a count value. and repeat the steps till your last day. Code is as below for your refrence i took monday as my intrested weekdays.

import datetime

A1=datetime.datetime.strptime("1/23/2016", "%m/%d/%Y")
A2=datetime.datetime.strptime("11/10/2016", "%m/%d/%Y")
count=0
week="Mon"
for i in range ((A2-A1).days):  #gives the no of days from A1 to A2
    if A1.strftime("%a")==week:
        count+=1
    A1+=datetime.timedelta(days=1)

print(count)

https://www.w3schools.com/python/python_datetime.asp

Upvotes: 0

Priyom
Priyom

Reputation: 41

Number of Mondays in 2020 can be got using numpy library

    import numpy as np
    np.busday_count('2020', '2021', weekmask='Mon')

Upvotes: 4

Arne
Arne

Reputation: 1338

If anyone need an even simpler answer,

from datetime import  date

d1 = date(2017, 1, 4)
d2 = date(2017, 1, 31)

count = 0

for d_ord in range(d1.toordinal(), d2.toordinal()):
    d = date.fromordinal(d_ord)
    if (d.weekday() == 4):
        count += 1

print(count)

Upvotes: 0

blubberdiblub
blubberdiblub

Reputation: 4135

This is efficient - even in the face of ten thousands of days between start and end - and still very flexible (it iterates at most 7 times inside the sum function):

def intervening_weekdays(start, end, inclusive=True, weekdays=[0, 1, 2, 3, 4]):
    if isinstance(start, datetime.datetime):
        start = start.date()               # make a date from a datetime

    if isinstance(end, datetime.datetime):
        end = end.date()                   # make a date from a datetime

    if end < start:
        # you can opt to return 0 or swap the dates around instead
        raise ValueError("start date must be before end date")

    if inclusive:
        end += datetime.timedelta(days=1)  # correct for inclusivity

    try:
        # collapse duplicate weekdays
        weekdays = {weekday % 7 for weekday in weekdays}
    except TypeError:
        weekdays = [weekdays % 7]

    ref = datetime.date.today()                    # choose a reference date
    ref -= datetime.timedelta(days=ref.weekday())  # and normalize its weekday

    # sum up all selected weekdays (max 7 iterations)
    return sum((ref_plus - start).days // 7 - (ref_plus - end).days // 7
               for ref_plus in
               (ref + datetime.timedelta(days=weekday) for weekday in weekdays))

This takes both datetime.date as well as datetime.datetime objects for start and end, respectively.

Also, you can choose between a closed (inclusive=True) and a half-open (inclusive=False) interval.

By default, it calculates the number of workdays between the dates, but you can choose any set of weekdays (weekend days: weekdays=[5, 6]) or single weekdays (Wednesdays: weekdays=2) as well.

Upvotes: 2

T.Okahara
T.Okahara

Reputation: 1234

This code still uses a for loop and an if/else. 

import datetime
import calendar

def weekday_count(start, end):
  start_date  = datetime.datetime.strptime(start, '%d/%m/%Y')
  end_date    = datetime.datetime.strptime(end, '%d/%m/%Y')
  week        = {}
  for i in range((end_date - start_date).days):
    day       = calendar.day_name[(start_date + datetime.timedelta(days=i+1)).weekday()]
    week[day] = week[day] + 1 if day in week else 1
  return week

print(weekday_count("01/01/2017", "31/01/2017"))

# prints result
# {'Monday': 5, 'Tuesday': 5, 'Friday': 4, 'Wednesday': 4, 'Thursday': 4, 'Sunday': 5, 'Saturday': 4}

Upvotes: 6

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