Generate list of months between interval in python
With pandas, you can have a one liner like this:
import pandas as pd
date1 = "2014-10-10" # input start date
date2 = "2016-01-07" # input end date
month_list = [i.strftime("%b-%y") for i in pd.date_range(start=date1, end=date2, freq='MS')]
>>> from datetime import datetime, timedelta
>>> from collections import OrderedDict
>>> dates = ["2014-10-10", "2016-01-07"]
>>> start, end = [datetime.strptime(_, "%Y-%m-%d") for _ in dates]
>>> OrderedDict(((start + timedelta(_)).strftime(r"%b-%y"), None) for _ in xrange((end - start).days)).keys()
['Oct-14', 'Nov-14', 'Dec-14', 'Jan-15', 'Feb-15', 'Mar-15', 'Apr-15', 'May-15', 'Jun-15', 'Jul-15', 'Aug-15', 'Sep-15', 'Oct-15', 'Nov-15', 'Dec-15', 'Jan-16']
Update: a bit of explanation, as requested in one comment. There are three problems here: parsing the dates into appropriate data structures (strptime
); getting the date range given the two extremes and the step (one month); formatting the output dates (strftime
). The datetime
type overloads the subtraction operator, so that end - start
makes sense. The result is a timedelta
object that represents the difference between the two dates, and the .days
attribute gets this difference expressed in days. There is no .months
attribute, so we iterate one day at a time and convert the dates to the desired output format. This yields a lot of duplicates, which the OrderedDict
removes while keeping the items in the right order.
Now this is simple and concise because it lets the datetime module do all the work, but it's also horribly inefficient. We're calling a lot of methods for each day while we only need to output months. If performance is not an issue, the above code will be just fine. Otherwise, we'll have to work a bit more. Let's compare the above implementation with a more efficient one:
from datetime import datetime, timedelta
from collections import OrderedDict
dates = ["2014-10-10", "2016-01-07"]
def monthlist_short(dates):
start, end = [datetime.strptime(_, "%Y-%m-%d") for _ in dates]
return OrderedDict(((start + timedelta(_)).strftime(r"%b-%y"), None) for _ in xrange((end - start).days)).keys()
def monthlist_fast(dates):
start, end = [datetime.strptime(_, "%Y-%m-%d") for _ in dates]
total_months = lambda dt: dt.month + 12 * dt.year
mlist = []
for tot_m in xrange(total_months(start)-1, total_months(end)):
y, m = divmod(tot_m, 12)
mlist.append(datetime(y, m+1, 1).strftime("%b-%y"))
return mlist
assert monthlist_fast(dates) == monthlist_short(dates)
if __name__ == "__main__":
from timeit import Timer
for func in "monthlist_short", "monthlist_fast":
print func, Timer("%s(dates)" % func, "from __main__ import dates, %s" % func).timeit(1000)
On my laptop, I get the following output:
monthlist_short 2.3209939003
monthlist_fast 0.0774540901184
The concise implementation is about 30 times slower, so I would not recommend it in time-critical applications :)
I found a very succinct way to do this with Pandas, sharing in case it helps anybody:
UPDATE: I've got it down to a one-liner with the help of this post :)
pd.date_range('2014-10-10','2016-01-07',
freq='MS').strftime("%Y-%b").tolist()
OLD ANSWER:
daterange = pd.date_range('2014-10-10','2016-01-07' , freq='1M')
daterange = daterange.union([daterange[-1] + 1])
daterange = [d.strftime('%y-%b') for d in daterange]
The second line prevents the last date from getting clipped off the list.