Generating all dates within a given range in python

Question

I have two string variables which contain dates in yyyy-mm-dd format as follows :

date1 = \'2011-05-03\'
date2 = \'2011-05-10\'

I want to write

Answer 1

Assuming your dates are already as a datetime.date class you can use .fromordinal and .toordinal to create this oneliner.

from datetime import date

start_date = date(2011, 5, 3)
end_date = date(2011, 5, 10)

[date.fromordinal(i) for i in range(start_date.toordinal(), end_date.toordinal())]

The result is exclusive end_date. Use end_date.toordinal() + 1 for a range inclusive end_date.

Answer 2

I like this one because it is intuitive and it gives an array of date strings.

import re
import datetime

def datetime_to_str_date(dt):
    return re.sub(r'\T.+$','', dt.isoformat())

start_date = datetime.datetime.strptime('2016-01-01', '%Y-%m-%d')
end_date = datetime.datetime.today()
num_of_days = (end_date - start_date).days

date_list = map(
        datetime_to_str_date, 
        [start_date + datetime.timedelta(days=x) for x in range(0, num_of_days)]
)

Answer 3

import time

def dates_between(start, end):
  start_epoch = int(time.mktime(time.strptime(start, "%Y-%m-%d")))
  end_epoch = int(time.mktime(time.strptime(end, "%Y-%m-%d"))) + 1 #include end
  return range(start_epoch, end_epoch, 86400)

Answer 4

import pandas as pd
date1 = '2011-05-03'
date2 = '2011-05-10'
pd.date_range(start = date1,end = date2)

Answer 5

Dates can be compared to each other just like numbers, and you can do date-related math with the datetime.timedelta object. There's no reason to use dateutil here, and there's no reason to hard-code the number of iterations a la 'range(9)'. This really becomes similar to how you'd deal with plain old numbers.

>>> import datetime
>>> date1 = '2011-05-03'
>>> date2 = '2011-05-10'
>>> start = datetime.datetime.strptime(date1, '%Y-%m-%d')
>>> end = datetime.datetime.strptime(date2, '%Y-%m-%d')
>>> step = datetime.timedelta(days=1)
>>> while start <= end:
...     print start.date()
...     start += step
... 
2011-05-03
2011-05-04
2011-05-05
2011-05-06
2011-05-07
2011-05-08
2011-05-09
2011-05-10
>>> 

Answer 6

>>> for a in range(9):
...     print(datetime.date(2011, 05, 03) + datetime.timedelta(a))
...
2011-05-03
2011-05-04
2011-05-05
2011-05-06
2011-05-07
2011-05-08
2011-05-09
2011-05-10
2011-05-11

I'm not too sure whether the parsing of the strings was integral or just the way you started the question. If so, please disregard the answer as oversimplified