Reputation: 2113
I want to get all datetime
object of each day in the current month but I haven't done yet. The result expected I wanted:
[datetime.date(2014, 1, 1), datetime.date(2014, 1, 2),
datetime.date(2014, 1, 3), ..., datetime.date(2014, 1, 29),
datetime.date(2014, 1, 30), datetime.date(2014, 1, 31)]
How can I solve this issue?
Please show me your ideas or suggestions. Thanks!
Upvotes: 38
Views: 51144
Reputation: 342
from datetime import date, timedelta
def days_in_month(year, month):
return filter(lambda dt: dt.month == month, [date(year, month, 1) + timedelta(days=i) for i in range(31)])
Upvotes: 1
Reputation: 174
With this solution, you will able to get all datetime object of each day in the current month.
import datetime, calendar
today = datetime.datetime.now()
month = today.month
year = today.year
days_per_month = calendar.monthrange(year, month)[1]
days_list = [datetime.date(year, month, day) for day in range(1, days_per_month + 1)]
Upvotes: 0
Reputation: 121
Ran across this and wanted to carry along timezone.
def get_month_days(date: datetime):
counter = datetime(date.year, date.month, 1, tzinfo=date.tzinfo)
date_list = []
while date.month == counter.month:
date_list.append(counter)
counter += timedelta(days=1)
return date_list
Upvotes: 2
Reputation: 47
Just in case:
import pandas as pd
print(pd.date_range(start = '2019-4-1',end='2019-4-30', freq='D'))
We get :
DatetimeIndex(['2019-04-01', '2019-04-02', '2019-04-03', '2019-04-04',
'2019-04-05', '2019-04-06', '2019-04-07', '2019-04-08',
'2019-04-09', '2019-04-10', '2019-04-11', '2019-04-12',
'2019-04-13', '2019-04-14', '2019-04-15', '2019-04-16',
'2019-04-17', '2019-04-18', '2019-04-19', '2019-04-20',
'2019-04-21', '2019-04-22', '2019-04-23', '2019-04-24',
'2019-04-25', '2019-04-26', '2019-04-27', '2019-04-28',
'2019-04-29', '2019-04-30'],
dtype='datetime64[ns]', freq='D')
Upvotes: 2
Reputation: 36171
There are several methods, but with calendar.monthrange
, you can get a tuple (first_weekday, nb_days)
:
>>> from calendar import monthrange
>>> monthrange(2014, 2)
(5, 28)
>>> a = monthrange(2014, 2)
>>> range(1, a[1]+1)
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28]
>>> [datetime.date(2014, 2, day) for day in range(1, a[1]+1)]
[datetime.date(2014, 1, 1), datetime.date(2014, 1, 2), datetime.date(2014, 1, 3), datetime.date(2014, 1, 4), datetime.date(2014, 1, 5), datetime.date(2014, 1, 6), datetime.date(2014, 1, 7), datetime.date(2014, 1, 8), datetime.date(2014, 1, 9), datetime.date(2014, 1, 10), datetime.date(2014, 1, 11), datetime.date(2014, 1, 12), datetime.date(2014, 1, 13), datetime.date(2014, 1, 14), datetime.date(2014, 1, 15), datetime.date(2014, 1, 16), datetime.date(2014, 1, 17), datetime.date(2014, 1, 18), datetime.date(2014, 1, 19), datetime.date(2014, 1, 20), datetime.date(2014, 1, 21), datetime.date(2014, 1, 22), datetime.date(2014, 1, 23), datetime.date(2014, 1, 24), datetime.date(2014, 1, 25), datetime.date(2014, 1, 26), datetime.date(2014, 1, 27), datetime.date(2014, 1, 28), datetime.date(2014, 1, 29), datetime.date(2014, 1, 30), datetime.date(2014, 1, 31)]
With a function to make it cleaner:
from calendar import monthrange
import datetime
def get_datetime_range(year, month):
nb_days = monthrange(year, month)[1]
return [datetime.date(year, month, day) for day in range(1, nb_days+1)]
print get_datetime_range(2014, 2) # usage example
Upvotes: 5
Reputation: 2191
>>> from calendar import Calendar
>>> Calendar().monthdayscalendar(2014,1)
[[0, 0, 1, 2, 3, 4, 5], [6, 7, 8, 9, 10, 11, 12], [13, 14, 15, 16, 17, 18, 19], [20, 21, 22, 23, 24, 25, 26], [27, 28, 29, 30, 31, 0, 0]]
>>> month = Calendar().itermonthdates(2014,1)
>>> print [day for day in month if day.month == 1] # this function will pad to make up full weeks, so filter it by month
[datetime.date(2014, 1, 1), datetime.date(2014, 1, 2), datetime.date(2014, 1, 3), datetime.date(2014, 1, 4), datetime.date(2014, 1, 5), datetime.date(2014, 1, 6), datetime.date(2014, 1, 7), datetime.date(2014, 1, 8), datetime.date(2014, 1, 9), datetime.date(2014, 1, 10), datetime.date(2014, 1, 11), datetime.date(2014, 1, 12), datetime.date(2014, 1, 13), datetime.date(2014, 1, 14), datetime.date(2014, 1, 15), datetime.date(2014, 1, 16), datetime.date(2014, 1, 17), datetime.date(2014, 1, 18), datetime.date(2014, 1, 19), datetime.date(2014, 1, 20), datetime.date(2014, 1, 21), datetime.date(2014, 1, 22), datetime.date(2014, 1, 23), datetime.date(2014, 1, 24), datetime.date(2014, 1, 25), datetime.date(2014, 1, 26), datetime.date(2014, 1, 27), datetime.date(2014, 1, 28), datetime.date(2014, 1, 29), datetime.date(2014, 1, 30), datetime.date(2014, 1, 31)]
http://docs.python.org/3/library/calendar.html
Upvotes: 3
Reputation: 10360
Here's a solution with datetime
and calendar
:
>>> import datetime, calendar
>>> year = 2014
>>> month = 1
>>> num_days = calendar.monthrange(year, month)[1]
>>> days = [datetime.date(year, month, day) for day in range(1, num_days+1)]
>>> days
[datetime.date(2014, 1, 1), datetime.date(2014, 1, 2), datetime.date(2014, 1, 3), datetime.date(2014, 1, 4), datetime.date(2014, 1, 5), datetime.date(2014, 1, 6), datetime.date(2014, 1, 7), datetime.date(2014, 1, 8), datetime.date(2014, 1, 9), datetime.date(2014, 1, 10), datetime.date(2014, 1, 11), datetime.date(2014, 1, 12), datetime.date(2014, 1, 13), datetime.date(2014, 1, 14), datetime.date(2014, 1, 15), datetime.date(2014, 1, 16), datetime.date(2014, 1, 17), datetime.date(2014, 1, 18), datetime.date(2014, 1, 19), datetime.date(2014, 1, 20), datetime.date(2014, 1, 21), datetime.date(2014, 1, 22), datetime.date(2014, 1, 23), datetime.date(2014, 1, 24), datetime.date(2014, 1, 25), datetime.date(2014, 1, 26), datetime.date(2014, 1, 27), datetime.date(2014, 1, 28), datetime.date(2014, 1, 29), datetime.date(2014, 1, 30), datetime.date(2014, 1, 31)]
Upvotes: 66
Reputation: 8062
Using datetime
and timedelta
:
import datetime
d0 = datetime.datetime(year=2014, month=1, day=1)
d1 = datetime.datetime(year=2014, month=2, day=1)
print (d1 - d0).days
>>> 31
Upvotes: 3