Python - Iterate thru month dates and print a custom output

Question

I'm currently unsure on the logic to be used for the below problem and new to programming as well.(Currently learning python)

Trying to iterate thru every date for a given month - say 05/01 -- 05/31 and print it out in the below format.

Monday thru Friday dates are to be printed separately. Saturday & Sunday dates are to be printed separately.

If the month starts on say Friday - 05/01/2020, ouput should be like as, its the last weekday of that week.

For the month of April 2020, output would be like below, as April month's 1st week started on Wednesday.

I managed to comeup with the below try, but not sure how to proceed further.

import sys
from datetime import date, datetime, timedelta

year = int(sys.argv[1])
month = int(sys.argv[2])
st_dt = int(sys.argv[3])
en_dt = int(sys.argv[4])

first_date = datetime(year, month, st_dt).date()
get_first_day = datetime(year, month, st_dt).isoweekday()

def daterange(startDate, endDate, delta=timedelta(days=1)):
    currentDate = startDate
    while currentDate <= endDate:
        yield currentDate
        currentDate += delta

for date in daterange(date(year, month, st_dt), date(year, month, en_dt), delta=timedelta(days=1)):
     print(date)


  date.py 2020 5 1 31 # script

Came up with a standalone 'if loop' and as i said before, not sure how to construct the bigger picture :(

if get_first_day == 1:
        #print("Monday")
        sec_d =  first_date + timedelta(days=4)
elif get_first_day == 2:
        sec_d = first_date + timedelta(days=3)
elif get_first_day == 3:
        sec_d = first_date + timedelta(days=2)
elif get_first_day == 4:
        sec_d = first_date + timedelta(days=2)
elif get_first_day == 5:
        sec_d = first_date
        #print("Friday")
else:
        pass

print(f"Second date:{sec_d} ") -- which gave  -- > Second date:2020-05-01

Answer 1

You could keep the dates in a dictionary, dictionary key is tuple of calendar week and type of day (weekend, day of the week).

Each day is saved by in the allDays dictionary, grouped by the combination of weeknum and type of day as key:

 ('18', 'weekend'): [datetime.date(2020, 5, 2), datetime.date(2020, 5, 3)],
 ('18', 'working'): [datetime.date(2020, 5, 1)],
 ('19', 'weekend'): [datetime.date(2020, 5, 9), datetime.date(2020, 5, 10)],
 ('19', 'working'): [datetime.date(2020, 5, 4), ...

So you just need to take out the fist and last item of each dict item:

import sys
from datetime import date, datetime, timedelta

year, month, st_dt, en_dt = 2020, 5, 1, 31

first_date = datetime(year, month, st_dt).date()
get_first_day = datetime(year, month, st_dt).isoweekday()

def daterange(startDate, endDate, delta=timedelta(days=1)):
    currentDate = startDate
    while currentDate <= endDate:
        yield currentDate
        currentDate += delta

allDays = {}
_lastDayType = None
for dte in daterange(date(year, month, st_dt), date(year, month, en_dt), delta=timedelta(days=1)):
    if 0 <= dte.weekday() < 5:
        _dayType = 'working'
    else:
        _dayType = 'weekend'

    _weeknum = dte.strftime("%V")  # number of calendar week
    _key = (_weeknum, _dayType)
    if _key not in allDays:        # create an empty list if unique key doesnt exist
        allDays[_key] = []
    allDays[_key].append(dte)      # add the dates ...

for k,v in allDays.items():
    if len(v) == 1:
        first, last = v[0], v[0]
    else:
        first, last = v[0], v[-1]
    print("%s >> %s" % (first, last))

Output:

2020-05-01 >> 2020-05-01
2020-05-02 >> 2020-05-03
2020-05-04 >> 2020-05-08
2020-05-09 >> 2020-05-10
2020-05-11 >> 2020-05-15
2020-05-16 >> 2020-05-17
2020-05-18 >> 2020-05-22
2020-05-23 >> 2020-05-24
2020-05-25 >> 2020-05-29
2020-05-30 >> 2020-05-31