I have the following code to do this, but how can I do it better? Right now I think it\'s better than nested loops, but it starts to get Perl-one-linerish when you have a ge
Can't* believe this question has existed for 9 years without anyone suggesting a simple recursive function:
from datetime import datetime, timedelta
def walk_days(start_date, end_date):
if start_date <= end_date:
print(start_date.strftime("%Y-%m-%d"))
next_date = start_date + timedelta(days=1)
walk_days(next_date, end_date)
#demo
start_date = datetime(2009, 5, 30)
end_date = datetime(2009, 6, 9)
walk_days(start_date, end_date)
Output:
2009-05-30
2009-05-31
2009-06-01
2009-06-02
2009-06-03
2009-06-04
2009-06-05
2009-06-06
2009-06-07
2009-06-08
2009-06-09
Edit: *Now I can believe it -- see Does Python optimize tail recursion? . Thank you Tim.
Here's code for a general date range function, similar to Ber's answer, but more flexible:
def count_timedelta(delta, step, seconds_in_interval):
"""Helper function for iterate. Finds the number of intervals in the timedelta."""
return int(delta.total_seconds() / (seconds_in_interval * step))
def range_dt(start, end, step=1, interval='day'):
"""Iterate over datetimes or dates, similar to builtin range."""
intervals = functools.partial(count_timedelta, (end - start), step)
if interval == 'week':
for i in range(intervals(3600 * 24 * 7)):
yield start + datetime.timedelta(weeks=i) * step
elif interval == 'day':
for i in range(intervals(3600 * 24)):
yield start + datetime.timedelta(days=i) * step
elif interval == 'hour':
for i in range(intervals(3600)):
yield start + datetime.timedelta(hours=i) * step
elif interval == 'minute':
for i in range(intervals(60)):
yield start + datetime.timedelta(minutes=i) * step
elif interval == 'second':
for i in range(intervals(1)):
yield start + datetime.timedelta(seconds=i) * step
elif interval == 'millisecond':
for i in range(intervals(1 / 1000)):
yield start + datetime.timedelta(milliseconds=i) * step
elif interval == 'microsecond':
for i in range(intervals(1e-6)):
yield start + datetime.timedelta(microseconds=i) * step
else:
raise AttributeError("Interval must be 'week', 'day', 'hour' 'second', \
'microsecond' or 'millisecond'.")
from datetime import date,timedelta
delta = timedelta(days=1)
start = date(2020,1,1)
end=date(2020,9,1)
loop_date = start
while loop_date<=end:
print(loop_date)
loop_date+=delta
Show the last n days from today:
import datetime
for i in range(0, 100):
print((datetime.date.today() + datetime.timedelta(i)).isoformat())
Output:
2016-06-29
2016-06-30
2016-07-01
2016-07-02
2016-07-03
2016-07-04
Why not try:
import datetime as dt
start_date = dt.datetime(2012, 12,1)
end_date = dt.datetime(2012, 12,5)
total_days = (end_date - start_date).days + 1 #inclusive 5 days
for day_number in range(total_days):
current_date = (start_date + dt.timedelta(days = day_number)).date()
print current_date
I have a similar problem, but I need to iterate monthly instead of daily.
This is my solution
import calendar
from datetime import datetime, timedelta
def days_in_month(dt):
return calendar.monthrange(dt.year, dt.month)[1]
def monthly_range(dt_start, dt_end):
forward = dt_end >= dt_start
finish = False
dt = dt_start
while not finish:
yield dt.date()
if forward:
days = days_in_month(dt)
dt = dt + timedelta(days=days)
finish = dt > dt_end
else:
_tmp_dt = dt.replace(day=1) - timedelta(days=1)
dt = (_tmp_dt.replace(day=dt.day))
finish = dt < dt_end
Example #1
date_start = datetime(2016, 6, 1)
date_end = datetime(2017, 1, 1)
for p in monthly_range(date_start, date_end):
print(p)
Output
2016-06-01
2016-07-01
2016-08-01
2016-09-01
2016-10-01
2016-11-01
2016-12-01
2017-01-01
Example #2
date_start = datetime(2017, 1, 1)
date_end = datetime(2016, 6, 1)
for p in monthly_range(date_start, date_end):
print(p)
Output
2017-01-01
2016-12-01
2016-11-01
2016-10-01
2016-09-01
2016-08-01
2016-07-01
2016-06-01