I'm trying to replace the day in my if statement for my date but I keep getting this output for my year.
05/15/5 besides 05/15/2020 . Code is below:
today_date = datetime.datetime.now()
date = today_date.date()
formatted_date = datetime.date.strftime(date, "%m/%d/%Y")
mmonth = date.month
myear = date.year
mdate = date.day
if mdate < 7:
m0weekend = formatted_date.replace(str(myear),str(mmonth),1)
else:
m0weekend = formatted_date.replace(str(myear),str(mmonth),15)
it's easier to replace the day before converting to a string:
date = date.replace(day=1)
or, in your case:
if mdate < 7:
m0weekend = date.replace(day=1)
else:
m0weekend = date.replace(day=15)
formatted_date is actually a string.
You are using the str.replace() method not the datetime.date.replace() method.
import datetime
today_date = datetime.datetime.now()
pre_formatted_date = today_date.date()
mmonth = pre_formatted_date.month
myear = pre_formatted_date.year
mdate = pre_formatted_date.day
if mdate < 7:
pre_formatted_date = pre_formatted_date.replace(day=1)
else:
pre_formatted_date = pre_formatted_date.replace(day=15)
print(pre_formatted_date)
formatted_date = pre_formatted_date.strftime("%m/%d/%Y")
print(formatted_date)
Which has the following output:
2020-05-15
05/15/2020
You might get today datetime.date directly from datetime rather than creating datetime.datetime and converting to date. After you have today you might create needed datetime.date and turn it into str, i.e.:
import datetime
today = datetime.date.today()
date = datetime.date(today.year, today.month, 1 if today.day < 7 else 15)
formatted_date = datetime.date.strftime(date, "%m/%d/%Y")
print(formatted_date) # 05/15/2020
Related
Can I get some help to create a list of days and months in the folowing format:
collection = ['2108', '2109', '2110', '2111', '2112', '2201']
I am trying in this way:
def make_collections(start_date: Date, end_date: Date):
result = []
date = start_date
while True:
if start_date >= end_date:
return result
date = date.strftime('%y%m%d%H%M%S')
result.append(date[:4])
date = (date.strptime(date, '%y%m%d%H%M%S')) + timedelta(days=1)
# test = MakeDataFrame()
# test.run()
if __name__ == '__main__':
start = datetime.now() - timedelta(days=365)
print(make_collections(start, datetime.now()))
But it doesn't work.
I want to give a start date and end date as an argument in function and make a list as I mentioned above with year and month.
Can I get some help to make a a simple function with start and end date as an arguments?
Thanks
I revised your code a little bit, so that the code will print what you expected. I used dateutil.relativedelta module because timedelta provide only day-based delta. dateutil.relativedelta support to calculate month and year differences.
from datetime import datetime, timedelta
from dateutil.relativedelta import relativedelta
def make_collections(start_date, end_date):
result = []
while start_date <= end_date:
result.append(str(start_date.year)[2:] + str(start_date.month).zfill(2))
start_date += relativedelta(months=1)
return result
if __name__ == '__main__':
start = datetime.now() - timedelta(days=365)
print(make_collections(start, datetime.now()))
#['2101', '2102', '2103', '2104', '2105', '2106', '2107', '2108', '2109', '2110', '2111', '2112', '2201']
python get the quarterly dates from a date range
example :
start date = 01/04/2020
end date = 01/04/2021
Here I need to get the Quaternary dates from this date range.
Try:
start_date = "01/04/2020"
end_date = "01/04/2021"
pd.date_range(start_date, end_date, freq='Q')
DatetimeIndex(['2020-03-31', '2020-06-30', '2020-09-30', '2020-12-31'], dtype='datetime64[ns]', freq='Q-DEC')
pd.date_range(start date, end date, freq='3m').to_period('Q')
With pure Python:
import datetime
start_date_str = "01/04/2020"
end_date_str = "01/04/2021"
start_date = datetime.datetime.strptime(start_date_str, "%d/%m/%Y").date()
end_date = datetime.datetime.strptime(end_date_str, "%d/%m/%Y").date()
print(f"Quarters within {start_date_str} and {end_date_str}:")
start_of_quarter = start_date
while True:
far_future = start_of_quarter + datetime.timedelta(days=93)
start_of_next_quarter = far_future.replace(day=1)
end_of_quarter = start_of_next_quarter - datetime.timedelta(days=1)
if end_of_quarter > end_date:
break
print(f"\t{start_of_quarter:%d/%m/%Y} - {end_of_quarter:%d/%m/%Y}")
start_of_quarter = start_of_next_quarter
Language Python
I am wondering if anyone can help me print out some dates.
i cam trying to create a loop in which i pass in a date say 01/1/2017 and the loop will then output the first and last day in every month between then and the present day.
Example
01/01/2017
31/01/2017
01/02/2017
28/02/2017
etc
Any help will be appreciated
Hope this will help,
Code:
from datetime import date
from dateutil.relativedelta import relativedelta
from calendar import monthrange
d1 = date(2018, 2, 26)
d2 = date.today()
def print_month_day_range(date):
first_day = date.replace(day = 1)
last_day = date.replace(day = monthrange(date.year, date.month)[1])
print (first_day.strftime("%d/%m/%Y"))
print (last_day.strftime("%d/%m/%Y"))
print_month_day_range(d1)
while (d1 < d2):
d1 = d1 + relativedelta(months=1)
print_month_day_range(d1)
Output:
01/02/2018
28/02/2018
01/03/2018
31/03/2018
...
01/07/2018
31/07/2018
you can use calendar module. Below is the code:
import datetime
from calendar import monthrange
strt_date = '01/1/2017'
iter_year = datetime.datetime.strptime(strt_date, '%m/%d/%Y').year
iter_month = datetime.datetime.strptime(strt_date, '%m/%d/%Y').month
cur_year = datetime.datetime.today().year
cur_month = datetime.datetime.today().month
while cur_year > iter_year or (cur_year == iter_year and iter_month <= cur_month):
number_of_days_in_month = monthrange(iter_year, iter_month)[1]
print '/'.join([str(iter_month) , '01', str(iter_year)])
print '/'.join([str(iter_month), str(number_of_days_in_month), str(iter_year)])
if iter_month == 12:
iter_year += 1
iter_month = 1
else:
iter_month += 1
this could work, as long as the first date you give is always the first of the month
from datetime import datetime
from datetime import timedelta
date_string = '01/01/2017'
date = datetime.strptime(date_string, '%d/%m/%Y').date()
today = datetime.now().date()
months = range(1,13)
years = range(date.year, today.year + 1)
for y in years:
for m in months:
new_date = date.replace(month=m, year=y)
last_day = new_date - timedelta(days=1)
if (date < new_date) & (new_date <= today):
print last_day.strftime('%d/%m/%Y')
print new_date.strftime('%d/%m/%Y')
elif (date <= new_date) & (new_date <= today):
print new_date.strftime('%d/%m/%Y')
This code would print the first and last days of all months in a year.
Maybe add some logic to iterate through the years
import datetime
def first_day_of_month(any_day):
first_month = any_day.replace(day=1)
return first_month
def last_day_of_month(any_day):
next_month = any_day.replace(day=28) + datetime.timedelta(days=4)
return next_month - datetime.timedelta(days=next_month.day)
for month in range(1, 13):
print first_day_of_month(datetime.date(2017, month, 1))
print last_day_of_month(datetime.date(2017, month, 1))
I want to count summer days between two dates. Summer is May first to August last.
This will count all days:
import datetime
startdate=datetime.datetime(2015,1,1)
enddate=datetime.datetime(2016,6,1)
delta=enddate-startdate
print delta.days
>>517
But how can only count the passed summer days?
You could define a generator to iterate over every date between startdate and enddate, define a function to check if a date represents a summer day and use sum to count the summer days:
import datetime
startdate = datetime.datetime(2015,1,1)
enddate = datetime.datetime(2016,6,1)
all_dates = (startdate + datetime.timedelta(days=x) for x in range(0, (enddate-startdate).days))
def is_summer_day(date):
return 5 <= date.month <= 8
print(sum(1 for date in all_dates if is_summer_day(date)))
# 154
Thanks to the generator, you don't need to create a huge list in memory with every day between startdate and enddate.
This iteration still considers every single day, even if it's not needed. For very large gaps, you could use the fact that every complete year has 123 summer days according to your definition.
You can create a few functions to count how many summer days you have between two days:
from datetime import date
def get_summer_start(year):
return date(year, 5, 1)
def get_summer_end(year):
return date(year, 8, 31)
def get_start_date(date, year):
return max(date, get_summer_start(year))
def get_end_date(date, year):
return min(date, get_summer_end(year))
def count_summer_days(date1, date2):
date1_year = date1.year
date2_year = date2.year
if date1_year == date2_year:
s = get_start_date(date1, date1_year)
e = get_end_date(date2, date1_year)
return (e - s).days
else:
s1 = max(date1, get_summer_start(date1_year))
e1 = get_summer_end(date1_year)
first_year = max(0,(e1 -s1).days)
s1 = get_summer_start(date2_year)
e1 = min(date2, get_summer_end(date2_year))
last_year = max(0,(e2 -s2).days)
other_years = date2_year - date1_year - 1
summer_days_per_year = (get_summer_end(date1_year) - get_summer_start(date1_year)).days
return first_year + last_year + (other_years * summer_days_per_year)
date1 = date(2015,1,1)
date2 = date(2016,6,1)
print count_summer_days(date1, date2)
Here is a better solution for large periods:
first_summer_day = (5,1)
last_summer_day = (8,31)
from datetime import date
startdate = date(2015,1,1)
enddate = date(2016,6,1)
# make sure that startdate > endate
if startdate > enddate:
startdate, endate = endate, startdate
def iter_yearly_summer_days(startdate, enddate):
for year in range(startdate.year, enddate.year+1):
start_period = startdate if year == startdate.year else date(year, 1, 1)
end_period = enddate if year == enddate.year else date(year, 12, 31)
year_first_summer_day = date(year, *first_summer_day)
year_last_summer_day = date(year, *last_summer_day)
summer_days_that_year = (min(year_last_summer_day, end_period) - max(year_first_summer_day, start_period)).days
print('year {} had {} days of summer'.format(year, summer_days_that_year))
yield summer_days_that_year
print(sum(iter_yearly_summer_days(startdate, enddate)))
I want to create a module can do this : inputdate = "27 Feb 2017", nomonths = 2, so outputdate must be "27 April 2017". inputdate and nomonths are 2 fields we have to fill in, and ouputdate is just a "read-only" field in odoo.
class DateGenerate(models.Model):
_name = "studentmanagement.dategenerate"
inputdate = fields.Date()
nomonths = fields.Integer(required=True)
outputdate = fields.Date(readonly=True)
#api.onchange('inputdate','nomonths')
def add_month(self):
for record in self:
dt = fields.Datetime.to_string(record.inputdate)
inpYear = datetime.strptime(dt,"%Y")
inpMonth = datetime.strptime(dt,"%m")
inpDay = datetime.strptime(dt,"%d")
outYear = inpYear + int((inpMonth + record.nomonths - 1)/12)
outMonth = (inpMonth + record.nomonths - 1) % 12 + 1
record.outputdate = datetime.date(outMonth, inpDay, outYear)
I write the code based on the sources and explanations on the internet, but it does not work and causes the error :
I read the code and direction from these links
https://docs.python.org/2/library/datetime.html
https://github.com/odoo/odoo/blob/10.0/odoo/fields.py#L1504
To add/take months onto a date/datetime object I would suggest using the python package: dateutil.relativedelta. http://dateutil.readthedocs.io/en/stable/relativedelta.html
basic usage:
from datetime import date, datetime
from dateutil.relativedelta import relativedelta
print date.today() + relativedelta(months=+3)
print datetime.today() + relativedelta(months=+3)
print date.today() - relativedelta(months=+3)
print datetime.today() - relativedelta(months=+3)
Your question references both datetime and date objects so have included examples for both.
Hope this helps.