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Iterating through a range of dates in Python


Question

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 generator in a list comprehension.

day_count = (end_date - start_date).days + 1
for single_date in [d for d in (start_date + timedelta(n) for n in range(day_count)) if d <= end_date]:
    print strftime("%Y-%m-%d", single_date.timetuple())

Notes

  • I'm not actually using this to print. That's just for demo purposes.
  • The start_date and end_date variables are datetime.date objects because I don't need the timestamps. (They're going to be used to generate a report).

Sample Output

For a start date of 2009-05-30 and an end date of 2009-06-09:

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
2013/01/15
1
385
1/15/2013 12:10:19 AM

Accepted Answer

Why are there two nested iterations? For me it produces the same list of data with only one iteration:

for single_date in (start_date + timedelta(n) for n in range(day_count)):
    print ...

And no list gets stored, only one generator is iterated over. Also the "if" in the generator seems to be unnecessary.

After all, a linear sequence should only require one iterator, not two.

Update after discussion with John Machin:

Maybe the most elegant solution is using a generator function to completely hide/abstract the iteration over the range of dates:

from datetime import timedelta, date

def daterange(start_date, end_date):
    for n in range(int((end_date - start_date).days)):
        yield start_date + timedelta(n)

start_date = date(2013, 1, 1)
end_date = date(2015, 6, 2)
for single_date in daterange(start_date, end_date):
    print(single_date.strftime("%Y-%m-%d"))

NB: For consistency with the built-in range() function this iteration stops before reaching the end_date. So for inclusive iteration use the next day, as you would with range().

2020/06/26
574
6/26/2020 12:02:15 PM


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.

2012/09/10

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)
2014/07/23

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.

2009/06/29

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
2016/10/13

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
2014/05/23

Source: https://stackoverflow.com/questions/1060279
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