Iterating through a range of dates in Python

This might be more clear:

from datetime import date, timedelta

start_date = date(2019, 1, 1)
end_date = date(2020, 1, 1)
delta = timedelta(days=1)
while start_date <= end_date:
    print(start_date.strftime("%Y-%m-%d"))
    start_date += delta

Use the dateutil library:

from datetime import date
from dateutil.rrule import rrule, DAILY

a = date(2009, 5, 30)
b = date(2009, 6, 9)

for dt in rrule(DAILY, dtstart=a, until=b):
    print dt.strftime("%Y-%m-%d")

This python library has many more advanced features, some very useful, like relative deltas—and is implemented as a single file (module) that's easily included into a project.

Pandas is great for time series in general, and has direct support for date ranges.

import pandas as pd
daterange = pd.date_range(start_date, end_date)

You can then loop over the daterange to print the date:

for single_date in daterange:
    print (single_date.strftime("%Y-%m-%d"))

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 http://pandas.pydata.org/pandas-docs/stable/timeseries.html#generating-ranges-of-timestamps

The power of Pandas is really its dataframes, which support vectorized operations (much like numpy) that make operations across large quantities of data very fast and easy.

EDIT: You could also completely skip the for loop and just print it directly, which is easier and more efficient:

print(daterange)

This is the most human-readable solution I can think of.

import datetime

def daterange(start, end, step=datetime.timedelta(1)):
    curr = start
    while curr < end:
        yield curr
        curr += step

import datetime

def daterange(start, stop, step=datetime.timedelta(days=1), inclusive=False):
  # inclusive=False to behave like range by default
  if step.days > 0:
    while start < stop:
      yield start
      start = start + step
      # not +=! don't modify object passed in if it's mutable
      # since this function is not restricted to
      # only types from datetime module
  elif step.days < 0:
    while start > stop:
      yield start
      start = start + step
  if inclusive and start == stop:
    yield start

# ...

for date in daterange(start_date, end_date, inclusive=True):
  print strftime("%Y-%m-%d", date.timetuple())

This function does more than you strictly require, by supporting negative step, etc. As long as you factor out your range logic, then you don't need the separate day_count and most importantly the code becomes easier to read as you call the function from multiple places.

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

Numpy's arange function can be applied to dates:

import numpy as np
from datetime import datetime, timedelta
d0 = datetime(2009, 1,1)
d1 = datetime(2010, 1,1)
dt = timedelta(days = 1)
dates = np.arange(d0, d1, dt).astype(datetime)

The use of astype is to convert from numpy.datetime64 to an array of datetime.datetime objects.

For completeness, Pandas also has a period_range function for timestamps that are out of bounds:

import pandas as pd

pd.period_range(start='1/1/1626', end='1/08/1627', freq='D')

Show the next 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
...

import datetime

def daterange(start, stop, step_days=1):
    current = start
    step = datetime.timedelta(step_days)
    if step_days > 0:
        while current < stop:
            yield current
            current += step
    elif step_days < 0:
        while current > stop:
            yield current
            current += step
    else:
        raise ValueError("daterange() step_days argument must not be zero")

if __name__ == "__main__":
    from pprint import pprint as pp
    lo = datetime.date(2008, 12, 27)
    hi = datetime.date(2009, 1, 5)
    pp(list(daterange(lo, hi)))
    pp(list(daterange(hi, lo, -1)))
    pp(list(daterange(lo, hi, 7)))
    pp(list(daterange(hi, lo, -7))) 
    assert not list(daterange(lo, hi, -1))
    assert not list(daterange(hi, lo))
    assert not list(daterange(lo, hi, -7))
    assert not list(daterange(hi, lo, 7)) 

for i in range(16):
    print datetime.date.today() + datetime.timedelta(days=i)

You can generate a series of date between two dates using the pandas library simply and trustfully

import pandas as pd

print pd.date_range(start='1/1/2010', end='1/08/2018', freq='M')

You can change the frequency of generating dates by setting freq as D, M, Q, Y (daily, monthly, quarterly, yearly )

import datetime
from dateutil.rrule import DAILY,rrule

date=datetime.datetime(2019,1,10)

date1=datetime.datetime(2019,2,2)

for i in rrule(DAILY , dtstart=date,until=date1):
     print(i.strftime('%Y%b%d'),sep='\n')

OUTPUT:

2019Jan10
2019Jan11
2019Jan12
2019Jan13
2019Jan14
2019Jan15
2019Jan16
2019Jan17
2019Jan18
2019Jan19
2019Jan20
2019Jan21
2019Jan22
2019Jan23
2019Jan24
2019Jan25
2019Jan26
2019Jan27
2019Jan28
2019Jan29
2019Jan30
2019Jan31
2019Feb01
2019Feb02

Using pendulum.period:

import pendulum

start = pendulum.from_format('2020-05-01', 'YYYY-MM-DD', formatter='alternative')
end = pendulum.from_format('2020-05-02', 'YYYY-MM-DD', formatter='alternative')

period = pendulum.period(start, end)

for dt in period:
    print(dt.to_date_string())

For those who are interested in Pythonic functional way:

from datetime import date, timedelta
from itertools import count, takewhile

for d in takewhile(lambda x: x<=date(2009,6,9), map(lambda x:date(2009,5,30)+timedelta(days=x), count())):
    print(d)

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

> pip install DateTimeRange

from datetimerange import DateTimeRange

def dateRange(start, end, step):
        rangeList = []
        time_range = DateTimeRange(start, end)
        for value in time_range.range(datetime.timedelta(days=step)):
            rangeList.append(value.strftime('%m/%d/%Y'))
        return rangeList

    dateRange("2018-09-07", "2018-12-25", 7)  

    Out[92]: 
    ['09/07/2018',
     '09/14/2018',
     '09/21/2018',
     '09/28/2018',
     '10/05/2018',
     '10/12/2018',
     '10/19/2018',
     '10/26/2018',
     '11/02/2018',
     '11/09/2018',
     '11/16/2018',
     '11/23/2018',
     '11/30/2018',
     '12/07/2018',
     '12/14/2018',
     '12/21/2018']

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

You can use Arrow:

This is example from the docs, iterating over hours:

from arrow import Arrow

>>> start = datetime(2013, 5, 5, 12, 30)
>>> end = datetime(2013, 5, 5, 17, 15)
>>> for r in Arrow.range('hour', start, end):
...     print repr(r)
...
<Arrow [2013-05-05T12:30:00+00:00]>
<Arrow [2013-05-05T13:30:00+00:00]>
<Arrow [2013-05-05T14:30:00+00:00]>
<Arrow [2013-05-05T15:30:00+00:00]>
<Arrow [2013-05-05T16:30:00+00:00]>

To iterate over days, you can use like this:

>>> start = Arrow(2013, 5, 5)
>>> end = Arrow(2013, 5, 5)
>>> for r in Arrow.range('day', start, end):
...     print repr(r)

(Didn't check if you can pass datetime.date objects, but anyways Arrow objects are easier in general)

If you are going to use dynamic timedelta then you can use:

1. With while loop

def datetime_range(start: datetime, end: datetime, delta: timedelta) -> Generator[datetime, None, None]:
    while start <= end:
        yield start
        start += delta

2. With for loop

from datetime import datetime, timedelta
from typing import Generator


def datetime_range(start: datetime, end: datetime, delta: timedelta) -> Generator[datetime, None, None]:
    delta_units = int((end - start) / delta)

    for _ in range(delta_units + 1):
        yield start
        start += delta

3. If you are using async/await

async def datetime_range(start: datetime, end: datetime, delta: timedelta) -> AsyncGenerator[datetime, None]:
    delta_units = int((end - start) / delta)

    for _ in range(delta_units + 1):
        yield start
        start += delta

4. List comprehension

def datetime_range(start: datetime, end: datetime, delta: timedelta) -> List[datetime]:
    delta_units = int((end - start) / delta)
    return [start + (delta * index) for index in range(delta_units + 1)]

Then 1 and 2 solutions simply can be used like this

start = datetime(2020, 10, 10, 10, 00)
end = datetime(2022, 10, 10, 18, 00)
delta = timedelta(minutes=30)

result = [time_part for time_part in datetime_range(start, end, delta)]
# or 
for time_part in datetime_range(start, end, delta):
    print(time_part)

3-third solution can be used like this in async context. Because it retruns an async generator object, which can be used only in async context

start = datetime(2020, 10, 10, 10, 00)
end = datetime(2022, 10, 10, 18, 00)
delta = timedelta(minutes=30)

result = [time_part async for time_part in datetime_range(start, end, delta)]

async for time_part in datetime_range(start, end, delta):
    print(time_part)

The benefit of the solutions about is that all of them are using dynamic timedelta. This can be very usefull in cases when you do not know which time delta you will have.

What about the following for doing a range incremented by days:

for d in map( lambda x: startDate+datetime.timedelta(days=x), xrange( (stopDate-startDate).days ) ):
  # Do stuff here

• startDate and stopDate are datetime.date objects

For a generic version:

for d in map( lambda x: startTime+x*stepTime, xrange( (stopTime-startTime).total_seconds() / stepTime.total_seconds() ) ):
  # Do stuff here

• startTime and stopTime are datetime.date or datetime.datetime object (both should be the same type)
• stepTime is a timedelta object

Note that .total_seconds() is only supported after python 2.7 If you are stuck with an earlier version you can write your own function:

def total_seconds( td ):
  return float(td.microseconds + (td.seconds + td.days * 24 * 3600) * 10**6) / 10**6

This function has some extra features:

• can pass a string matching the DATE_FORMAT for start or end and it is converted to a date object
• can pass a date object for start or end

• error checking in case the end is older than the start

  import datetime
  from datetime import timedelta
  
  
  DATE_FORMAT = '%Y/%m/%d'
  
  def daterange(start, end):
        def convert(date):
              try:
                    date = datetime.datetime.strptime(date, DATE_FORMAT)
                    return date.date()
              except TypeError:
                    return date
  
        def get_date(n):
              return datetime.datetime.strftime(convert(start) + timedelta(days=n), DATE_FORMAT)
  
        days = (convert(end) - convert(start)).days
        if days <= 0:
              raise ValueError('The start date must be before the end date.')
        for n in range(0, days):
              yield get_date(n)
  
  
  start = '2014/12/1'
  end = '2014/12/31'
  print list(daterange(start, end))
  
  start_ = datetime.date.today()
  end = '2015/12/1'
  print list(daterange(start, end))
  

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'.")

in polars it can be done as well, if eager is set to True

start = datetime.date(year=2009, month=5, day=30)
end = datetime.date(year=2009, month=6, day=9)
dates = pl.date_range(start, end, eager=True)

for date in dates:
    print(date)

Slightly different approach to reversible steps by storing range args in a tuple.

def date_range(start, stop, step=1, inclusive=False):
    day_count = (stop - start).days
    if inclusive:
        day_count += 1

    if step > 0:
        range_args = (0, day_count, step)
    elif step < 0:
        range_args = (day_count - 1, -1, step)
    else:
        raise ValueError("date_range(): step arg must be non-zero")

    for i in range(*range_args):
        yield start + timedelta(days=i)

Here is an alternative solution based on leftjoin solution.

For pendulum below 3.0.0

def test_iteration():
    start = pendulum.from_format('2018-01', 'YYYY-MM')
    end = pendulum.from_format('2020-01', 'YYYY-MM')

    interval = pendulum.period(start, end)

    for dt in interval.range('months'):
        print(dt.format('YYYY-MM'))

And for pendulum 3.0.0 and above (they have renamed period to interval) https://github.com/sdispater/pendulum/pull/676

def test_iteration():
    start = pendulum.from_format('2018-01', 'YYYY-MM')
    end = pendulum.from_format('2020-01', 'YYYY-MM')

    interval = pendulum.interval(start, end)

    for dt in interval.range('months'):
        print(dt.format('YYYY-MM'))

You can change de range values from the list on the documentation here : https://pendulum.eustace.io/docs/#range

Supported units for range() are: years, months, weeks, days, hours, minutes, seconds and microseconds

Lots of answers here. All of them are terrible. :-D

from datetime import date
for o in range(start.toordinal(), end.toordinal() + 1):
    print(date.fromordinal(o))

to/from ordinal converts dates to and from monotonic integers.

Life is easier when you know how to live.

Like a lot of solutions here, this works for input values that are dates or datetimes (which are a subclass).

The more_itertools package provides the numeric_range function, which allows the following:

from more_itertools import numeric_range

for single_date in numeric_range(start_date, end_date, timedelta(days=1)):
    ...

Answer for Python API Call Loop

using datetime and timedelta from the datetime module

from datetime import datetime, timedelta

import requests


headers = {"accept": "application/json", "x-api-key": "xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx"}

delta = timedelta(days=30)
start = datetime(year=2022, month=1, day=1)
num_ranges = 10

for _ in range(num_ranges):
    start_str = start.strftime("%Y-%m-%d")
    end = start + delta
    end_str = end.strftime("%Y-%m-%d")
    url = f"https://apiendpoint.com/v2/dapps/xxxx/history/uaw?dateFrom={start_str}&dateTo={end_str}"
    response = requests.request("GET", url, headers=headers)
    # TODO: append response json to result
    start = end