Print all thursdays between date range

Question

I want to print all Thursdays between these date ranges

from datetime import date, timedelta
sdate = date(2015, 1, 7)   # start date
edate = date(2015, 12, 31)   # end date

What is the best pythonic way to do that?

This is a duplicated question. It's answered here. https://stackoverflow.com/questions/42949449/using-pandas-date-range-to-generate-multiple-datetimes-two-dates-per-week — ERIC, Aug 11 '21 at 11:45

score 2 · Answer 1 · answered Jun 08 '21 at 07:43

using your sdate.weekday() # returns int between 0 (mon) and 6 (sun):

sdate = ...
while sdate < edate:
    if sdate.weekday() != 3:  # not thursday
        sdate += timedelta(days=1)
        continue
    # It is thursday
    print(sdate)
    sdate += timedelta(days=7)  #  next week

Whole Brain · Answer 2 · 2021-06-08T09:05:11.400

2

Compute the number of days till thursday from the start date :

days_to_thursday = (3 - sdate.weekday()) % 7

Compute the number of thursdays inbetween both dates:

week_diff = ((edate - sdate).days - days_to_thursday ) // 7

Get all thursdays in the date range:

thursdays = [sdate + timedelta(days=days_to_thursday + 7 * more_weeks) \
    for more_weeks in range(week_diff + 1) ]

Print them if you need:

for t in thursdays:
    print(t)

edited Jun 08 '21 at 09:05

answered Jun 08 '21 at 08:05

Whole Brain

2,097
2
8
18

1

Thank you for pointing out the trick of the remainder of dividing a negative number by a positive number. I didn't know it before, very useful property. – Алексей Р Jun 08 '21 at 09:50

Nitish Gupta · Answer 3 · 2021-06-08T08:04:13.373

1

import datetime
import calendar

def weekday_count(start, end, day):
    start_date  = datetime.datetime.strptime(start, '%d/%m/%Y')
    end_date    = datetime.datetime.strptime(end, '%d/%m/%Y')
    day_count = []
    
    for i in range((end_date - start_date).days):
        if calendar.day_name[(start_date + datetime.timedelta(days=i+1)).weekday()] == day:
            print(str(start_date + datetime.timedelta(days=i+1)).split()[0])

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

# prints result
# 2017-01-05
# 2017-01-12
# 2017-01-19
# 2017-01-26

edited Jun 08 '21 at 08:04

answered Jun 08 '21 at 07:39

Nitish Gupta

39
6

I want to print thurdays between date range – sam Jun 08 '21 at 07:40
I dont need to print total Thursdays but dates of thursdays – sam Jun 08 '21 at 07:52
what if I dont want time in that and only need to print date – sam Jun 08 '21 at 07:57
You can try to extract date from deltatime object or convert it to str followed by split and indexing (shown in the edited answer) – Nitish Gupta Jun 08 '21 at 08:07

sobolevn · Answer 4 · 2021-06-08T12:54:44.473

Simple solution:

from datetime import date, timedelta

sdate = date(2015, 1, 7)   # start date
edate = date(2015, 12, 31)   # end date

delta = edate - sdate

for day in range(delta.days + 1):
    day_obj = sdate + timedelta(days=day)
    if day_obj.weekday() == 3:  # Thursday
        print(day_obj)

# 2015-01-08
# 2015-01-15
# ...
# 2015-12-24
# 2015-12-31

The most efficient solution:

from datetime import date, timedelta

sdate = date(2015, 1, 7)   # start date
edate = date(2015, 12, 31)   # end date

day_index = 3  # Thursday
delta = (day_index - sdate.weekday()) % 7
match = sdate + timedelta(days=delta)

while match <= edate:  # Change this to `<` to ignore the last one
    print(match) # Can be easily converted to a generator with `yield`
    match += timedelta(days=7)

# 2015-01-08
# 2015-01-15
# ...
# 2015-12-24
# 2015-12-31

Docs:

.weekday(): https://docs.python.org/3/library/datetime.html#datetime.date.weekday
.timedelta(): https://docs.python.org/3/library/datetime.html#timedelta-objects
% operator on negative numbers: The modulo operation on negative numbers in Python

The `abs` function makes this solution incorrect **for any sdate with a weekday greater than 3**. It gives a delta of `1, 2 or 3` instead of `6, 5 or 4` respectively. — Whole Brain, Jun 08 '21 at 09:14
@sam looks like it does print all Thursdays for 2016: https://replit.com/@sobolevn/CarefulConcreteListener#main.py — sobolevn, Jun 08 '21 at 20:40

score 1 · Answer 5 · answered Jun 08 '21 at 08:02

You could try a list comprehension to get the Thursdays between the 2 dates.

This code actually outputs the dates as formatted strings but you can get actual dates by dropping the strftime.

from datetime import date, timedelta
sdate = date(2015, 1, 7)   # start date
edate = date(2015, 12, 31)   # end date

thursdays = [(sdate+timedelta(days=d)).strftime('%A %Y-%m-%d') for d in range(0, (edate-sdate).days+1) 
             if (sdate+timedelta(days=d)).weekday() ==3]

print('\n'.join(thursdays))

""" Example output
Thursday 2015-01-08
Thursday 2015-01-15
Thursday 2015-01-22
Thursday 2015-01-29
Thursday 2015-02-05
Thursday 2015-02-12
""""

score 1 · Accepted Answer · answered Jun 08 '21 at 08:53

Most people are iterating through every day which is a waste. Also might be helpful to delay calculating the thursdays until you actually need them. For this you could use a generator.

def get_days(start, day_index, end=None):
    # set the start as the next valid day
    start += timedelta(days=(day_index - start.weekday()) % 7)
    week = timedelta(days=7)
    while end and start < end or not end:
        yield start
        start += week

This delays getting the next day until you need it, and allows infinite days if you don't specify and end date.

thursday_generator = get_days(date(2015, 1, 7), 3, date(2015, 12, 31))
print(list(thursday_generator))

"""
[datetime.date(2015, 1, 8), datetime.date(2015, 1, 15), datetime.date(2015, 1, 22), ...]
"""

You can easily dump as strings:

print("\n".join(map(str, thursday_generator)))

"""
2015-01-08
2015-01-15
2015-01-22
...
"""

You can also use f-strings for custom string formatting:

print("\n".join(f"{day:%A %x}" for day in thursday_generator))

"""
Thursday 01/08/15
Thursday 01/15/15
Thursday 01/22/15
...
"""

If you don't specify an end date, it goes on forever.

In [28]: thursday_generator = get_days(date(2015, 1, 7), 3)
    ...: print(len(list(thursday_generator)))
---------------------------------------------------------------------------
OverflowError                             Traceback (most recent call last)
<ipython-input-28-b161cdcccc75> in <module>
      1 thursday_generator = get_days(date(2015, 1, 7), 3)
----> 2 print(len(list(thursday_generator)))

<ipython-input-16-0691db329606> in get_days(start, day_index, end)
      5     while end and start < end or not end:
      6         yield start
----> 7         start += week
      8

OverflowError: date value out of range

Print all thursdays between date range

6 Answers6