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Get the week numbers between two dates with python

I'd like to find the most pythonic way to output a list of the week numbers between two dates.
For example:
input
start = datetime.date(2011, 12, 25)
end = datetime.date(2012, 1, 21)
output
find_weeks(start, end)
>> [201152, 201201, 201202, 201203]
I've been struggling using the datetime library with little success

Something in the lines of (update: removed less-readable option)
import datetime
def find_weeks(start,end):
l = []
for i in range((end-start).days + 1):
d = (start+datetime.timedelta(days=i)).isocalendar()[:2] # e.g. (2011, 52)
yearweek = '{}{:02}'.format(*d) # e.g. "201152"
l.append(yearweek)
return sorted(set(l))
start = datetime.date(2011, 12, 25)
end = datetime.date(2012, 1, 21)
print(find_weeks(start,end)[1:]) # [1:] to exclude first week.
Returns
['201152', '201201', '201202', '201203']
To include the first week (201151) simply remove [1:] after function call

.isocalendar() is your friend here - it returns a tuple of (year, week of year, day of week). We use that to reset the start date to the start of th eweek, and then add on a week each time until we pass the end date:
import datetime
def find_weeks(start_date, end_date):
subtract_days = start_date.isocalendar()[2] - 1
current_date = start_date + datetime.timedelta(days=7-subtract_days)
weeks_between = []
while current_date <= end_date:
weeks_between.append(
'{}{:02d}'.format(*current_date.isocalendar()[:2])
)
current_date += datetime.timedelta(days=7)
return weeks_between
start = datetime.date(2011, 12, 25)
end = datetime.date(2012, 1, 21)
print(find_weeks(start, end))
This prints
['201152', '201201', '201202', '201203']

Using Pandas
import pandas as pd
dates=pd.date_range(start=start,end=end,freq='W')
date_index=dates.year.astype(str)+dates.weekofyear.astype(str).str.zfill(2)
date_index.tolist()

I suggest you the following easy-to-read solution:
import datetime
start = datetime.date(2011, 12, 25)
end = datetime.date(2012, 1, 21)
def find_weeks(start, end):
l = []
while (start.isocalendar()[1] != end.isocalendar()[1]) or (start.year != end.year):
l.append(start.isocalendar()[1] + 100*start.year)
start += datetime.timedelta(7)
l.append(start.isocalendar()[1] + 100*start.year)
return (l[1:])
print(find_weeks(start, end))
>> [201252, 201201, 201202, 201203]

I prefer the arrow style solution here (might need pip install arrow):
import arrow
start = arrow.get('2011-12-25')
end = arrow.get('2012-01-21')
weeks = list(arrow.Arrow.span_range('week', start, end))
result looks like this:
>> from pprint import pprint
>> pprint(weeks[1:])
[(<Arrow [2011-12-19T00:00:00+00:00]>,
<Arrow [2011-12-25T23:59:59.999999+00:00]>),
(<Arrow [2011-12-26T00:00:00+00:00]>,
<Arrow [2012-01-01T23:59:59.999999+00:00]>),
(<Arrow [2012-01-02T00:00:00+00:00]>,
<Arrow [2012-01-08T23:59:59.999999+00:00]>),
(<Arrow [2012-01-09T00:00:00+00:00]>,
<Arrow [2012-01-15T23:59:59.999999+00:00]>),
(<Arrow [2012-01-16T00:00:00+00:00]>,
<Arrow [2012-01-22T23:59:59.999999+00:00]>)]
from there you can change the output to match the year and week number.

How to find number of Fridays/Thursdays in date range

Is there any python function to deduce the number of Fridays or Thursdays from a date range? I searched google and I found many methods which usually use days divided by 7 concept but it does not give you the accurate days. For example from 1/Nov/2016 to 12/Nov/2016 there are two Fridays and two Thursdays so the result of subtraction should be 8.

You can do it with numpy:
import numpy as np
from datetime import datetime
start_date = datetime(2022, 10, 19).strftime('%Y-%m-%d')
end_date = datetime(2022, 12, 21).strftime('%Y-%m-%d')
weekend_days = np.busday_count(start_date, end_date, weekmask='0000110').item()
numpy busday_count doc
Keep in mind that end date is excluded from the count.

Using the date object from the datetime module.
from datetime import date, timedelta
curr = date(2016, 11, 1)
end = date(2016, 11, 12)
step = timedelta(1)
num_thur_fri = 0
while curr <= end:
if curr.weekday() in [3,2]: #Friday and thursday
num_thur_fri += 1
curr += step
print(num_thur_fri)
More reading here: https://docs.python.org/2/library/datetime.html#module-datetime
#brianpck is right, this is a really naive solution. Here's a better one
from datetime import date
begin = date(2016, 11, 1)
end = date(2016, 11, 12)
diff = (begin-end).days
day_of_week = begin.weekday()
num_thur_fri = 2*(diff//7)
for i in range(diff%7):
if day_of_week in [2,3]:
num_thur_fri += 1
day_of_week = (day_of_week +1) %7

Here is a simpler and faster approach that will calculate this figure for long periods of time.
First, you must calculate the amount of days between two datetime's. You can then floor divide by 7 to get the amount of entire weeks and multiply by 2 to get the number of Thursdays and Fridays. The final step is to modulo by seven to get the amount of days at the tail and then calculate how many of those are Thursdays or Fridays: this last step is the only one that actually requires knowing which weekday it is.
A full function would be:
from datetime import datetime, timedelta
def thursday_fridays_between(date1, date2):
days_between = abs((date2 - date1).days)
thursday_friday = days_between // 7 * 2
thursday_friday += sum((a + timedelta(i)).weekday() in (3, 2) for i in range(days_between % 7 + 1))
return thursday_friday
It can be used as follows:
>>> a = datetime(2016, 11, 1)
>>> b = datetime(2016, 11, 12)
>>> thursday_fridays_between(a, b)
4

i figure out a method, correct me if i am wrong.
here is my code
from datetime import date, timedelta, datetime
curr = "1-11-2016"
end = "30-11-2016"
format = "%d-%m-%Y"
start_date = datetime.strptime(curr, format)
end_date = datetime.strptime(end, format)
step = timedelta(1)
num_thur_fri = 0
off_days = ['Fri','Thu']
days = (end_date - start_date).days
for x in range(days):
day = start_date.strftime("%a")
print(day)
if day in off_days:
num_thur_fri += 1
start_date += step
print(num_thur_fri)

python - iteration by date and time - half hour

I think about how to iterate by data and time in following way:
2016-06-10,00:00
2016-06-10,01:30
2016-06-10,02:00
2016-06-10,02:30
....
Can somebody help me ?

Use timedelta to generate a range of datetime objects:
from datetime import timedelta, datetime
start_date = datetime(2016, 6, 9, 5, 0, 0)
for td in (start_date + timedelta(minutes=30*it) for it in xrange(10)):
print td.strftime("%Y-%m-%d,%H:%M")
Output:
2016-06-09,05:00
2016-06-09,05:30
2016-06-09,06:00
2016-06-09,06:30
2016-06-09,07:00
2016-06-09,07:30
2016-06-09,08:00
2016-06-09,08:30
2016-06-09,09:00
2016-06-09,09:30

import datetime
from datetime import timedelta
format = "%Y-%m-%d,%H:%M"
start = datetime.datetime(2016, 6, 10, 0, 0)
minutes = 0
for i in range(6):
print (start + timedelta(minutes=minutes)).strftime(format)
minutes += 30
Output:
$ python timespan.py
2016-06-10,00:00
2016-06-10,00:30
2016-06-10,01:00
2016-06-10,01:30
2016-06-10,02:00
2016-06-10,02:30

Python Remove From List of Times Based on Business Hours

I have a tuple list of times in 15 min intervals for 24h range. Now I need to remove times outside of business hours.
My times:
times = []
for i in range(0, 24 * 4):
times.append((datetime.combine(date.today(), time()) + timedelta(minutes=15) * i).time())
Business Hours:
time_open = '08:00:00'
time_close = '17:00:00'
How do I remove times from the list outside of the business hours?

Here's a quick and dirty solution if i understood what you want. You can just check if the time is between the hours the business is open / closed before appending to the list.
from datetime import datetime as d, timedelta, time
import datetime as o_d
time_open = '08:00:00'
time_close = '17:00:00'
times = []
for i in range(0, 24 * 4):
date = (d.combine(o_d.date.today(), time()) + timedelta(minutes=15) * i).time()
date_str = date.strftime("%H:%M:%S")
if time_open <= date_str <= time_close:
times.append(date_str)
#times.append(date)
print(times)
Edit from comments: If you wanted a tuple representation of what's typically called "military" time vs what most people are used to i.e. - AM / PM you can do the following
from datetime import datetime as d, timedelta, time
import datetime as o_d
time_open = '08:00:00'
time_close = '17:00:00'
times = []
for i in range(0, 24 * 4):
date = (d.combine(o_d.date.today(), time()) + timedelta(minutes=15) * i).time()
date_str = date.strftime("%H:%M:%S")
if time_open <= date_str <= time_close:
#If you want a list of just standard times
#times.append(date_str)
#If you want the time representation(s) here and no longer need the
#the datetime object
times.append((date_str, date.strftime("%I:%M %p"))
#If you still need the datetime object append this do list comprehension below in comment with #options = ....
#times.append(date)
print(times)
#assuming you have appended date above see comment
#options = [( c.strftime("%H:%M:%S"), c.strftime("%I:%M %p")) for c in times]

Creating a range of dates in Python

I want to create a list of dates, starting with today, and going back an arbitrary number of days, say, in my example 100 days. Is there a better way to do it than this?
import datetime
a = datetime.datetime.today()
numdays = 100
dateList = []
for x in range (0, numdays):
dateList.append(a - datetime.timedelta(days = x))
print dateList

Marginally better...
base = datetime.datetime.today()
date_list = [base - datetime.timedelta(days=x) for x in range(numdays)]

Pandas is great for time series in general, and has direct support for date ranges.
For example pd.date_range():
import pandas as pd
from datetime import datetime
datelist = pd.date_range(datetime.today(), periods=100).tolist()
It also has lots of options to make life easier. For example if you only wanted weekdays, you would just swap in bdate_range.
See date range documentation
In addition it fully supports pytz timezones and can smoothly span spring/autumn DST shifts.
EDIT by OP:
If you need actual python datetimes, as opposed to Pandas timestamps:
import pandas as pd
from datetime import datetime
pd.date_range(end = datetime.today(), periods = 100).to_pydatetime().tolist()
#OR
pd.date_range(start="2018-09-09",end="2020-02-02")
This uses the "end" parameter to match the original question, but if you want descending dates:
pd.date_range(datetime.today(), periods=100).to_pydatetime().tolist()

Get range of dates between specified start and end date (Optimized for time & space complexity):
import datetime
start = datetime.datetime.strptime("21-06-2014", "%d-%m-%Y")
end = datetime.datetime.strptime("07-07-2014", "%d-%m-%Y")
date_generated = [start + datetime.timedelta(days=x) for x in range(0, (end-start).days)]
for date in date_generated:
print date.strftime("%d-%m-%Y")

You can write a generator function that returns date objects starting from today:
import datetime
def date_generator():
from_date = datetime.datetime.today()
while True:
yield from_date
from_date = from_date - datetime.timedelta(days=1)
This generator returns dates starting from today and going backwards one day at a time. Here is how to take the first 3 dates:
>>> import itertools
>>> dates = itertools.islice(date_generator(), 3)
>>> list(dates)
[datetime.datetime(2009, 6, 14, 19, 12, 21, 703890), datetime.datetime(2009, 6, 13, 19, 12, 21, 703890), datetime.datetime(2009, 6, 12, 19, 12, 21, 703890)]
The advantage of this approach over a loop or list comprehension is that you can go back as many times as you want.
Edit
A more compact version using a generator expression instead of a function:
date_generator = (datetime.datetime.today() - datetime.timedelta(days=i) for i in itertools.count())
Usage:
>>> dates = itertools.islice(date_generator, 3)
>>> list(dates)
[datetime.datetime(2009, 6, 15, 1, 32, 37, 286765), datetime.datetime(2009, 6, 14, 1, 32, 37, 286836), datetime.datetime(2009, 6, 13, 1, 32, 37, 286859)]

yeah, reinvent the wheel....
just search the forum and you'll get something like this:
from dateutil import rrule
from datetime import datetime
list(rrule.rrule(rrule.DAILY,count=100,dtstart=datetime.now()))

You can also use the day ordinal to make it simpler:
def date_range(start_date, end_date):
for ordinal in range(start_date.toordinal(), end_date.toordinal()):
yield datetime.date.fromordinal(ordinal)
Or as suggested in the comments you can create a list like this:
date_range = [
datetime.date.fromordinal(ordinal)
for ordinal in range(
start_date.toordinal(),
end_date.toordinal(),
)
]

From the title of this question I was expecting to find something like range(), that would let me specify two dates and create a list with all the dates in between. That way one does not need to calculate the number of days between those two dates, if one does not know it beforehand.
So with the risk of being slightly off-topic, this one-liner does the job:
import datetime
start_date = datetime.date(2011, 1, 1)
end_date = datetime.date(2014, 1, 1)
dates_2011_2013 = [ start_date + datetime.timedelta(n) for n in range(int ((end_date - start_date).days))]
All credits to this answer!

Here's a slightly different answer building off of S.Lott's answer that gives a list of dates between two dates start and end. In the example below, from the start of 2017 to today.
start = datetime.datetime(2017,1,1)
end = datetime.datetime.today()
daterange = [start + datetime.timedelta(days=x) for x in range(0, (end-start).days)]

If there are two dates and you need the range try
from dateutil import rrule, parser
date1 = '1995-01-01'
date2 = '1995-02-28'
datesx = list(rrule.rrule(rrule.DAILY, dtstart=parser.parse(date1), until=parser.parse(date2)))

Based on answers I wrote for myself this:
import datetime;
print [(datetime.date.today() - datetime.timedelta(days=x)).strftime('%Y-%m-%d') for x in range(-5, 0)]
Output:
['2017-12-11', '2017-12-10', '2017-12-09', '2017-12-08', '2017-12-07']
The difference is that I get the 'date' object, not the 'datetime.datetime' one.

A bit of a late answer I know, but I just had the same problem and decided that Python's internal range function was a bit lacking in this respect so I've overridden it in a util module of mine.
from __builtin__ import range as _range
from datetime import datetime, timedelta
def range(*args):
if len(args) != 3:
return _range(*args)
start, stop, step = args
if start < stop:
cmp = lambda a, b: a < b
inc = lambda a: a + step
else:
cmp = lambda a, b: a > b
inc = lambda a: a - step
output = [start]
while cmp(start, stop):
start = inc(start)
output.append(start)
return output
print range(datetime(2011, 5, 1), datetime(2011, 10, 1), timedelta(days=30))

Here is gist I created, from my own code, this might help. (I know the question is too old, but others can use it)
https://gist.github.com/2287345
(same thing below)
import datetime
from time import mktime
def convert_date_to_datetime(date_object):
date_tuple = date_object.timetuple()
date_timestamp = mktime(date_tuple)
return datetime.datetime.fromtimestamp(date_timestamp)
def date_range(how_many=7):
for x in range(0, how_many):
some_date = datetime.datetime.today() - datetime.timedelta(days=x)
some_datetime = convert_date_to_datetime(some_date.date())
yield some_datetime
def pick_two_dates(how_many=7):
a = b = convert_date_to_datetime(datetime.datetime.now().date())
for each_date in date_range(how_many):
b = a
a = each_date
if a == b:
continue
yield b, a

Here's a one liner for bash scripts to get a list of weekdays, this is python 3. Easily modified for whatever, the int at the end is the number of days in the past you want.
python -c "import sys,datetime; print('\n'.join([(datetime.datetime.today() - datetime.timedelta(days=x)).strftime(\"%Y/%m/%d\") for x in range(0,int(sys.argv[1])) if (datetime.datetime.today() - datetime.timedelta(days=x)).isoweekday()<6]))" 10
Here is a variant to provide a start (or rather, end) date
python -c "import sys,datetime; print('\n'.join([(datetime.datetime.strptime(sys.argv[1],\"%Y/%m/%d\") - datetime.timedelta(days=x)).strftime(\"%Y/%m/%d \") for x in range(0,int(sys.argv[2])) if (datetime.datetime.today() - datetime.timedelta(days=x)).isoweekday()<6]))" 2015/12/30 10
Here is a variant for arbitrary start and end dates. not that this isn't terribly efficient, but is good for putting in a for loop in a bash script:
python -c "import sys,datetime; print('\n'.join([(datetime.datetime.strptime(sys.argv[1],\"%Y/%m/%d\") + datetime.timedelta(days=x)).strftime(\"%Y/%m/%d\") for x in range(0,int((datetime.datetime.strptime(sys.argv[2], \"%Y/%m/%d\") - datetime.datetime.strptime(sys.argv[1], \"%Y/%m/%d\")).days)) if (datetime.datetime.strptime(sys.argv[1], \"%Y/%m/%d\") + datetime.timedelta(days=x)).isoweekday()<6]))" 2015/12/15 2015/12/30

A generic method that allows to create date ranges on parameterised window size(day, minute, hour, seconds):
from datetime import datetime, timedelta
def create_date_ranges(start, end, **interval):
start_ = start
while start_ < end:
end_ = start_ + timedelta(**interval)
yield (start_, min(end_, end))
start_ = end_
Tests:
def main():
tests = [
('2021-11-15:00:00:00', '2021-11-17:13:00:00', {'days': 1}),
('2021-11-15:00:00:00', '2021-11-16:13:00:00', {'hours': 12}),
('2021-11-15:00:00:00', '2021-11-15:01:45:00', {'minutes': 30}),
('2021-11-15:00:00:00', '2021-11-15:00:01:12', {'seconds': 30})
]
for t in tests:
print("\nInterval: %s, range(%s to %s)" % (t[2], t[0], t[1]))
start = datetime.strptime(t[0], '%Y-%m-%d:%H:%M:%S')
end = datetime.strptime(t[1], '%Y-%m-%d:%H:%M:%S')
ranges = list(create_date_ranges(start, end, **t[2]))
x = list(map(
lambda x: (x[0].strftime('%Y-%m-%d:%H:%M:%S'), x[1].strftime('%Y-%m-%d:%H:%M:%S')),
ranges
))
print(x)
main()
Test output:
Interval: {'days': 1}, range(2021-11-15:00:00:00 to 2021-11-17:13:00:00)
[('2021-11-15:00:00:00', '2021-11-16:00:00:00'), ('2021-11-16:00:00:00', '2021-11-17:00:00:00'), ('2021-11-17:00:00:00', '2021-11-17:13:00:00')]
Interval: {'hours': 12}, range(2021-11-15:00:00:00 to 2021-11-16:13:00:00)
[('2021-11-15:00:00:00', '2021-11-15:12:00:00'), ('2021-11-15:12:00:00', '2021-11-16:00:00:00'), ('2021-11-16:00:00:00', '2021-11-16:12:00:00'), ('2021-11-16:12:00:00', '2021-11-16:13:00:00')]
Interval: {'minutes': 30}, range(2021-11-15:00:00:00 to 2021-11-15:01:45:00)
[('2021-11-15:00:00:00', '2021-11-15:00:30:00'), ('2021-11-15:00:30:00', '2021-11-15:01:00:00'), ('2021-11-15:01:00:00', '2021-11-15:01:30:00'), ('2021-11-15:01:30:00', '2021-11-15:01:45:00')]
Interval: {'seconds': 30}, range(2021-11-15:00:00:00 to 2021-11-15:00:01:12)
[('2021-11-15:00:00:00', '2021-11-15:00:00:30'), ('2021-11-15:00:00:30', '2021-11-15:00:01:00'), ('2021-11-15:00:01:00', '2021-11-15:00:01:12')]

Matplotlib related
from matplotlib.dates import drange
import datetime
base = datetime.date.today()
end = base + datetime.timedelta(days=100)
delta = datetime.timedelta(days=1)
l = drange(base, end, delta)

I know this has been answered, but I'll put down my answer for historical purposes, and since I think it is straight forward.
import numpy as np
import datetime as dt
listOfDates=[date for date in np.arange(firstDate,lastDate,dt.timedelta(days=x))]
Sure it won't win anything like code-golf, but I think it is elegant.

Another example that counts forwards or backwards, starting from Sandeep's answer.
from datetime import date, datetime, timedelta
from typing import Sequence
def range_of_dates(start_of_range: date, end_of_range: date) -> Sequence[date]:
if start_of_range <= end_of_range:
return [
start_of_range + timedelta(days=x)
for x in range(0, (end_of_range - start_of_range).days + 1)
]
return [
start_of_range - timedelta(days=x)
for x in range(0, (start_of_range - end_of_range).days + 1)
]
start_of_range = datetime.today().date()
end_of_range = start_of_range + timedelta(days=3)
date_range = range_of_dates(start_of_range, end_of_range)
print(date_range)
gives
[datetime.date(2019, 12, 20), datetime.date(2019, 12, 21), datetime.date(2019, 12, 22), datetime.date(2019, 12, 23)]
and
start_of_range = datetime.today().date()
end_of_range = start_of_range - timedelta(days=3)
date_range = range_of_dates(start_of_range, end_of_range)
print(date_range)
gives
[datetime.date(2019, 12, 20), datetime.date(2019, 12, 19), datetime.date(2019, 12, 18), datetime.date(2019, 12, 17)]
Note that the start date is included in the return, so if you want four total dates, use timedelta(days=3)

from datetime import datetime , timedelta, timezone
start_date = '2022_01_25'
end_date = '2022_01_30'
start = datetime.strptime(start_date, "%Y_%m_%d")
print(type(start))
end = datetime.strptime(end_date, "%Y_%m_%d")
##pDate = str(pDate).replace('-', '_')
number_of_days = (end - start).days
print("number_of_days: ", number_of_days)
##
date_list = []
for day in range(number_of_days):
a_date = (start + timedelta(days = day)).astimezone(timezone.utc)
a_date = a_date.strftime('%Y-%m-%d')
date_list.append(a_date)
print(date_list)

A monthly date range generator with datetime and dateutil. Simple and easy to understand:
import datetime as dt
from dateutil.relativedelta import relativedelta
def month_range(start_date, n_months):
for m in range(n_months):
yield start_date + relativedelta(months=+m)

import datetime
def date_generator():
cur = base = datetime.date.today()
end = base + datetime.timedelta(days=100)
delta = datetime.timedelta(days=1)
while(end>base):
base = base+delta
print base
date_generator()

from datetime import datetime, timedelta
from dateutil import parser
def getDateRange(begin, end):
""" """
beginDate = parser.parse(begin)
endDate = parser.parse(end)
delta = endDate-beginDate
numdays = delta.days + 1
dayList = [datetime.strftime(beginDate + timedelta(days=x), '%Y%m%d') for x in range(0, numdays)]
return dayList

From above answers i created this example for date generator
import datetime
date = datetime.datetime.now()
time = date.time()
def date_generator(date, delta):
counter =0
date = date - datetime.timedelta(days=delta)
while counter <= delta:
yield date
date = date + datetime.timedelta(days=1)
counter +=1
for date in date_generator(date, 30):
if date.date() != datetime.datetime.now().date():
start_date = datetime.datetime.combine(date, datetime.time())
end_date = datetime.datetime.combine(date, datetime.time.max)
else:
start_date = datetime.datetime.combine(date, datetime.time())
end_date = datetime.datetime.combine(date, time)
print('start_date---->',start_date,'end_date---->',end_date)

I thought I'd throw in my two cents with a simple (and not complete) implementation of a date range:
from datetime import date, timedelta, datetime
class DateRange:
def __init__(self, start, end, step=timedelta(1)):
self.start = start
self.end = end
self.step = step
def __iter__(self):
start = self.start
step = self.step
end = self.end
n = int((end - start) / step)
d = start
for _ in range(n):
yield d
d += step
def __contains__(self, value):
return (
(self.start <= value < self.end) and
((value - self.start) % self.step == timedelta(0))
)