We Keep Coding

List iteration through JSON array with dates (date format conflict)

I've been diving into List Comprehensions, and I'm determined to put it into practice.
The below code takes a month and year input to determine the number of business days in the month, minus the public holidays (available at https://www.gov.uk/bank-holidays.json).
Additionally, I want to list all public holidays in that month/year, but I'm struck with a date format conflict.
TypeError: '<' not supported between instances of 'str' and 'datetime.date'
edate and sdate are datetime.date, whereas title["date"] is a string.
I've tried things like datetime.strptime and datetime.date to now avail.
How can I resolve the date conflict within the List Comprehension?
Any help or general feedback on code appreciated.
from datetime import date, timedelta, datetime
import inspect
from turtle import title
from typing import Iterator
import numpy as np
import json
import requests
from calendar import month, monthrange
import print
# Ask for a month and year input (set to September for quick testing)
monthInput = "09"
yearInput = "2022"
# Request from UK GOV and filter to England and Wales
holidaysJSON = requests.get("https://www.gov.uk/bank-holidays.json")
ukHolidaysJSON = json.loads(holidaysJSON.text)['england-and-wales']['events']
# List for all England and Wales holidays
ukHolidayList = []
eventIterator = 0
for events in ukHolidaysJSON:
ukHolidayDate = list(ukHolidaysJSON[eventIterator].values())[1]
ukHolidayList.append(ukHolidayDate)
eventIterator += 1
# Calculate days in the month
daysInMonth = monthrange(int(yearInput), int(monthInput))[1] # Extract the number of days in the month
# Define start and end dates
sdate = date(int(yearInput), int(monthInput), 1) # start date
edate = date(int(yearInput), int(monthInput), int(daysInMonth)) # end date
# Calculate delta
delta = edate - sdate
# Find all of the business days in the month
numberOfWorkingDays = 0
for i in range(delta.days + 1): # Look through all days in the month
day = sdate + timedelta(days=i)
if np.is_busday([day]) and str(day) not in ukHolidayList: # Determine if it's a business day
print("- " + str(day))
numberOfWorkingDays += 1
# Count all of the UK holidays
numberOfHolidays = 0
for i in range(delta.days + 1): # Look through all days in the month
day = sdate + timedelta(days=i)
if str(day) in ukHolidayList: # Determine if it's a uk holiday
numberOfHolidays += 1
# Strip the 0 from the month input
month = months[monthInput.lstrip('0')]
# for x in ukHolidaysJSON:
# pprint.pprint(x["title"])
# This is where I've gotten to
hols = [ title["title"] for title in ukHolidaysJSON if title["date"] < sdate and title["date"] > edate ]
print(hols)

I got this to work. You can used the datetime module to parse the string format but then you need to convert that result into a Date to compare to the Date objects you have already.
hols = [ title["title"] for title in ukHolidaysJSON if datetime.strptime(title["date"], '%Y-%m-%d').date() < sdate and datetime.strptime(title["date"], "%Y-%m-%d").date() > edate ]

First use strptime and then convert the datetime object to date. I'm not sure if there's a more straightforward way but this seems to work:
hols = [title["title"] for title in ukHolidaysJSON
if datetime.strptime(title["date"], "%Y-%M-%d").date() < sdate and
datetime.strptime(title["date"], "%Y-%M-%d").date() > edate]

How can I get the first day of the next month in Python?

How can I get the first date of the next month in Python? For example, if it's now 2019-12-31, the first day of the next month is 2020-01-01. If it's now 2019-08-01, the first day of the next month is 2019-09-01.
I came up with this:
import datetime
def first_day_of_next_month(dt):
'''Get the first day of the next month. Preserves the timezone.
Args:
dt (datetime.datetime): The current datetime
Returns:
datetime.datetime: The first day of the next month at 00:00:00.
'''
if dt.month == 12:
return datetime.datetime(year=dt.year+1,
month=1,
day=1,
tzinfo=dt.tzinfo)
else:
return datetime.datetime(year=dt.year,
month=dt.month+1,
day=1,
tzinfo=dt.tzinfo)
# Example usage (assuming that today is 2021-01-28):
first_day_of_next_month(datetime.datetime.now())
# Returns: datetime.datetime(2021, 2, 1, 0, 0)
Is it correct? Is there a better way?

Here is a 1-line solution using nothing more than the standard datetime library:
(dt.replace(day=1) + datetime.timedelta(days=32)).replace(day=1)
Examples:
>>> dt = datetime.datetime(2016, 2, 29)
>>> print((dt.replace(day=1) + datetime.timedelta(days=32)).replace(day=1))
2016-03-01 00:00:00
>>> dt = datetime.datetime(2019, 12, 31)
>>> print((dt.replace(day=1) + datetime.timedelta(days=32)).replace(day=1))
2020-01-01 00:00:00
>>> dt = datetime.datetime(2019, 12, 1)
>>> print((dt.replace(day=1) + datetime.timedelta(days=32)).replace(day=1))
2020-01-01 00:00:00

Using dateutil you can do it the most literally possible:
import datetime
from dateutil import relativedelta
today = datetime.date.today()
next_month = today + relativedelta.relativedelta(months=1, day=1)
In English: add 1 month(s) to the today's date and set the day (of the month) to 1. Note the usage of singular and plural forms of day(s) and month(s). Singular sets the attribute to a value, plural adds the number of periods.
You can store this relativedelta.relativedelta object to a variable and the pass it around. Other answers involve more programming logic.
EDIT You can do it with the standard datetime library as well, but it's not so beautiful:
next_month = (today.replace(day=1) + datetime.timedelta(days=32)).replace(day=1)
sets the date to the 1st of the current month, adds 32 days (or any number between 31 and 59 which guarantees to jump into the next month) and then sets the date to the 1st of that month.

you can use calendar to get the number of days in a given month, then add timedelta(days=...), like this:
from datetime import date, timedelta
from calendar import monthrange
days_in_month = lambda dt: monthrange(dt.year, dt.month)[1]
today = date.today()
first_day = today.replace(day=1) + timedelta(days_in_month(today))
print(first_day)
if you're fine with external deps, you can use dateutil (which I love...)
from datetime import date
from dateutil.relativedelta import relativedelta
today = date.today()
first_day = today.replace(day=1) + relativedelta(months=1)
print(first_day)

Extract the year and month, add 1 and form a new date using the year, month and day=1:
from datetime import date
now = date(2020,12,18)
y,m = divmod(now.year*12+now.month,12)
nextMonth = date(y,m+1,1)
print(now,nextMonth)
# 2020-12-18 2021-01-01

Your way looks good yet I would have done it this way:
import datetime
from dateutil import relativedelta
dt = datetime.datetime(year=1998,
month=12,
day=12)
nextmonth = dt + relativedelta.relativedelta(months=1)
nextmonth.replace(day=1)
print(nextmonth)

Using only python standard libraries:
import datetime
today = datetime.date.today()
first_of_next_month = return date.replace(
day=1,
month=date.month % 12 + 1,
year=date.year + (date.month // 12)
)
could be generalized to...
def get_first_of_month(date, month_offset=0):
# zero based indexing of month to make math work
month_count = date.month - 1 + month_offset
return date.replace(
day=1, month=month_count % 12 + 1, year=date.year + (month_count // 12)
)
first_of_next_month = get_first_of_month(today, 1)
Other solutions that don't require 3rd party libraries include:
Toby Petty's answer is another good option.
If the exact timedelta is helpful to you,
a slight modification on Adam.Er8's answer might be convenient:
import calendar, datetime
today = datetime.date.today()
time_until_next_month = datetime.timedelta(
calendar.monthrange(today.year, today.month)[1] - today.day + 1
)
first_of_next_month = today + time_until_next_month

With Zope's DateTime library a very simple solution is possible
from DateTime.DateTime import DateTime
date = DateTime() # today
while date.day() != 1:
date += 1
print(date)

I see so many wonderful solutions to this problem I personally was looking for a solution for getting the first and last day of the previous month when I stmbled on this question.
But here is a solution I like to think is quite simple and elegant:
date = datetime.datetime.now().date()
same_time_next_month = date + datetime.timedelta(days = date.day)
first_day_of_next_month_from_date = same_time_next_month - datetime.timedelta(days = same_time_next_month.day - 1)
Here we simply add the day of the target date to the date to get the same time of the next month, and then remove the number of days elapsed from the new date gotten.

Try this, for starting day of each month, change MonthEnd(1) to MonthBegin(1):
import pandas as pd
from pandas.tseries.offsets import MonthBegin, MonthEnd
date_list = (pd.date_range('2021-01-01', '2022-01-31',
freq='MS') + MonthEnd(1)).strftime('%Y-%m-%d').tolist()
date_list
Out:
['2021-01-31',
'2021-02-28',
'2021-03-31',
'2021-04-30',
'2021-05-31',
'2021-06-30',
'2021-07-31',
'2021-08-31',
'2021-09-30',
'2021-10-31',
'2021-11-30',
'2021-12-31',
'2022-01-31']

With python-dateutil:
from datetime import date
from dateutil.relativedelta import relativedelta
last day of current month:
date.today() + relativedelta(day=31)
first day of next month:
date.today() + relativedelta(day=31) + relativedelta(days=1)

How to get a specific date from the previous month given the current date in python?

I want to get the 20th of previous month, given the current_date()
I am trying to use time.strftime but not able to subtract the value from it.
timestr = time.strftime("%Y-(%m-1)%d")
This is giving me error. The expected output is 2019-03-20 if my current_date is in April. Not sure how to go about it.
I read the posts from SO and most of them address getting the first day / last day of the month. Any help would be appreciated.

from datetime import date, timedelta
today = date.today()
last_day_prev_month = today - timedelta(days=today.day)
twenty_prev_month = last_day_prev_month.replace(day=20)
print(twenty_prev_month) # 2019-03-20

Use datetime.replace
import datetime
current_date = datetime.date.today()
new_date = current_date.replace(
month = current_date.month - 1,
day = 20
)
print(new_date)
#2019-03-20
Edit
That won't work for Jan so this is a workaround:
import datetime
current_date = datetime.date(2019, 2, 17)
month = current_date.month - 1
year = current_date.year
if not month:
month, year = 12, year - 1
new_date = datetime.date(year=year, month=month, day=20)

I imagine it is the way dates are parsed. It is my understanding that with your code it is looking for
2019-(03-1)20 or 2019-(12-1)15, etc..
Because the %y is not a variable, but a message about how the date is to be expected within a string of text, and other characters are what should be expected, but not processed (like "-")
This seems entirely not what you are going for. I would just parse the date like normal and then reformat it to be a month earlier:
import datetime
time = datetime.datetime.today()
print(time)
timestr = time.strftime("%Y-%m-%d")
year, month, day = timestr.split("-")
print("{}-{}-{}".format(year, int(month)-1, day))
This would be easier with timedelta objects, but sadly there isn't one for months, because they are of various lengths.
To be more robust if a new year is involved:
import datetime
time = datetime.datetime.today()
print(time)
timestr = time.strftime("%Y-%m-%d")
year, month, day = timestr.split("-")
if month in [1, "01", "1"]: # I don't remember how January is represented
print("{}-{}-{}".format(int(year) - 1, 12, day)) # use December of last year
else:
print("{}-{}-{}".format(year, int(month)-1, day))

This will help:
from datetime import date, timedelta
dt = date.today() - timedelta(30)// timedelta(days No.)
print('Current Date :',date.today())
print(dt)

It is not possible to do math inside a string passed to time.strftime, but you can do something similar to what you're asking very easily using the time module
in Python 3
# Last month
t = time.gmtime()
print(f"{t.tm_year}-{t.tm_mon-1}-20")
or in Python 2
print("{0}-{1}-{2}".format(t.tm_year, t.tm_mon -1, 20))
If you have fewer constraints, you can just use the datetime module instead.

You could use datetime, dateutil or arrow to find the 20th day of the previous month. See examples below.
Using datetime:
from datetime import date
d = date.today()
month, year = (d.month-1, d.year) if d.month != 1 else (12, d.year-1)
last_month = d.replace(day=20, month=month, year=year)
print(last_month)
Using datetime and timedelta:
from datetime import date
from datetime import timedelta
d = date.today()
last_month = (d - timedelta(days=d.day)).replace(day=20)
print(last_month)
Using datetime and dateutil:
from datetime import date
from dateutil.relativedelta import relativedelta # pip install python-dateutil
d = date.today()
last_month = d.replace(day=20) - relativedelta(months=1)
print(last_month)
Using arrow:
import arrow # pip install arrow
d = arrow.now()
last_month = d.shift(months=-1).replace(day=20).datetime.date()
print(last_month)

Get last day of month in python

Getting error for December month.
ValueError: month must be in 1..12
def last_day_of_month(ds):
cur_ds = datetime.strptime(ds, '%Y-%m-%d')
next_month = datetime(year=cur_ds.year, month=cur_ds.month+1, day=1)
last_day_month = next_month - timedelta(days=1)
return datetime.strftime(last_day_month, '%Y-%m-%d')
print last_day_of_month('2016-12-01')

In line 3 month=cur_ds.month+1 you are giving 13th month which is not valid. If you want to calculate last day of a given month you could also use month range from calendar library.
>>import calendar
>>year, month = 2016, 12
>>calendar.monthrange(year, month)[1]
31

You can't make a datetime with a month of 13. So you have to find a way to fix it. A simple solution is to convert the incremented month to an extra year:
# Reduce 12 to 1, 0 and all other #s to 0, #
extrayear, month = divmod(cur_ds.month, 12)
# Add 1 or 0 to existing year, add one to month (which was reduced to 0-11)
next_month = datetime(year=cur_ds.year + extrayear, month=month + 1, day=1)

You're passing in 12 as current month, then adding one to get next_month, making it 13. Check for the 12 case and set month=1 instead.

this is how I did it.
from django.utils import timezone
from calendar import monthrange
from datetime import datetime
current = timezone.now()
firstdayofmonth = current.replace(day=1)
endmonth = monthrange(current.year, current.month)
lastdayofmonth = datetime(current.year, current.month, endmonth[1])

How to get the value of Python datetime

I want to get the value of datetime using python code
ex. 20141104
that is example what I want to get
then, How can I get the datetime like that.
import calendar
for month in range(1, 13):
year = 2014
make_calendar = calendar.monthcalendar(year, month)
for weekend in make_calendar:
for day in weekend:
if (day != 0):
parameter = str(year) + str(month) + str(day)
print parameter
-> I try to get value like example but, the result is 201442.
I want to 20140402 not 201442.
I'm in need of help.

user1153551 has shown how to do what you want using the calendar module, but you should consider using the datetime module instead, with its powerful strftime method. The calendar module is great when you need to manipulate and/or format calendar at the month or year level, but for lower level manipulation at the level of individual dates, datetime is probably more suitable.
For example:
#! /usr/bin/env python
from datetime import date, timedelta
#A timedelta object of 1 day
oneday = timedelta(days=1)
year = 2014
#A date object of the start of the year
current_day = date(year, 1, 1)
#Print all the days of the given year in YYYYmmdd format
while current_day.year == year:
print current_day.strftime("%Y%m%d")
current_day += oneday

You can use following code to get desired output:
from time import gmtime, strftime,time, sleep
date = strftime("%Y%m%d")
print date

Use '%02d' % month to archive day, month with leading zero for Example,
>>> import datetime
>>> '%02d' % datetime.date.today().month
'11'
Python Code
import calendar
for month in range(1, 13):
year = 2014
make_calendar = calendar.monthcalendar(year, month)
for weekend in make_calendar:
for day in weekend:
if (day != 0):
parameter = '%02d%02d%02d' % (year, month, day)
print parameter