I have two different dates and I want to know the difference in days between them. The format of the date is YYYY-MM-DD.

I have a function that can ADD or SUBTRACT a given number to a date:

def addonDays(a, x):
   ret = time.strftime("%Y-%m-%d",time.localtime(time.mktime(time.strptime(a,"%Y-%m-%d"))+x*3600*24+3600))      
   return ret

where A is the date and x the number of days I want to add. And the result is another date.

I need a function where I can give two dates and the result would be an int with date difference in days.

Use - to get the difference between two datetime objects and take the days member.

from datetime import datetime

def days_between(d1, d2):
    d1 = datetime.strptime(d1, "%Y-%m-%d")
    d2 = datetime.strptime(d2, "%Y-%m-%d")
    return abs((d2 - d1).days)

Another short solution:

from datetime import date

def diff_dates(date1, date2):
    return abs(date2-date1).days

def main():
    d1 = date(2013,1,1)
    d2 = date(2013,9,13)
    result1 = diff_dates(d2, d1)
    print '{} days between {} and {}'.format(result1, d1, d2)
    print ("Happy programmer's day!")

main()

I tried the code posted by larsmans above but, there are a couple of problems:

1) The code as is will throw the error as mentioned by mauguerra 2) If you change the code to the following:

...
    d1 = d1.strftime("%Y-%m-%d")
    d2 = d2.strftime("%Y-%m-%d")
    return abs((d2 - d1).days)

This will convert your datetime objects to strings but, two things

1) Trying to do d2 – d1 will fail as you cannot use the minus operator on strings and 2) If you read the first line of the above answer it stated, you want to use the – operator on two datetime objects but, you just converted them to strings

What I found is that you literally only need the following:

import datetime

end_date = datetime.datetime.utcnow()
start_date = end_date - datetime.timedelta(days=8)
difference_in_days = abs((end_date - start_date).days)

print difference_in_days

Try this:

data=pd.read_csv('C:\Users\Desktop\Data Exploration.csv')
data.head(5)
first=data['1st Gift']
last=data['Last Gift']
maxi=data['Largest Gift']
l_1=np.mean(first)-3*np.std(first)
u_1=np.mean(first)+3*np.std(first)


m=np.abs(data['1st Gift']-np.mean(data['1st Gift']))>3*np.std(data['1st Gift'])
pd.value_counts(m)
l=first[m]
data.loc[:,'1st Gift'][m==True]=np.mean(data['1st Gift'])+3*np.std(data['1st Gift'])
data['1st Gift'].head()




m=np.abs(data['Last Gift']-np.mean(data['Last Gift']))>3*np.std(data['Last Gift'])
pd.value_counts(m)
l=last[m]
data.loc[:,'Last Gift'][m==True]=np.mean(data['Last Gift'])+3*np.std(data['Last Gift'])
data['Last Gift'].head()

pd.date_range(‘2019-01-01’, ‘2019-02-01’).shape[0]

Sure I could write this myself, but before I go reinventing the wheel is there a function that already does this?

Given an instance x of datetime.date, (x.month-1)//3 will give you the quarter (0 for first quarter, 1 for second quarter, etc — add 1 if you need to count from 1 instead;-).

Originally two answers, multiply upvoted and even originally accepted (both currently deleted), were buggy — not doing the -1 before the division, and dividing by 4 instead of 3. Since .month goes 1 to 12, it’s easy to check for yourself what formula is right:

for m in range(1, 13):
  print m//4 + 1,
print

gives 1 1 1 2 2 2 2 3 3 3 3 4 — two four-month quarters and a single-month one (eep).

for m in range(1, 13):
  print (m-1)//3 + 1,
print

gives 1 1 1 2 2 2 3 3 3 4 4 4 — now doesn’t this look vastly preferable to you?-)

This proves that the question is well warranted, I think;-).

I don’t think the datetime module should necessarily have every possible useful calendric function, but I do know I maintain a (well-tested;-) datetools module for the use of my (and others’) projects at work, which has many little functions to perform all of these calendric computations — some are complex, some simple, but there’s no reason to do the work over and over (even simple work) or risk bugs in such computations;-).

IF you are already using pandas, it’s quite simple.

import datetime as dt
import pandas as pd

quarter = pd.Timestamp(dt.date(2016, 2, 29)).quarter
assert quarter == 1

If you have a date column in a dataframe, you can easily create a new quarter column:

df['quarter'] = df['date'].dt.quarter

I would suggest another arguably cleaner solution. If X is a datetime.datetime.now() instance, then the quarter is:

import math
Q=math.ceil(X.month/3.)

ceil has to be imported from math module as it can’t be accessed directly.

For anyone trying to get the quarter of the fiscal year, which may differ from the calendar year, I wrote a Python module to do just this.

Installation is simple. Just run:

$ pip install fiscalyear

There are no dependencies, and fiscalyear should work for both Python 2 and 3.

It’s basically a wrapper around the built-in datetime module, so any datetime commands you are already familiar with will work. Here’s a demo:

>>> from fiscalyear import *
>>> a = FiscalDate.today()
>>> a
FiscalDate(2017, 5, 6)
>>> a.fiscal_year
2017
>>> a.quarter
3
>>> b = FiscalYear(2017)
>>> b.start
FiscalDateTime(2016, 10, 1, 0, 0)
>>> b.end
FiscalDateTime(2017, 9, 30, 23, 59, 59)
>>> b.q3
FiscalQuarter(2017, 3)
>>> b.q3.start
FiscalDateTime(2017, 4, 1, 0, 0)
>>> b.q3.end
FiscalDateTime(2017, 6, 30, 23, 59, 59)

fiscalyear is hosted on GitHub and PyPI. Documentation can be found at Read the Docs. If you’re looking for any features that it doesn’t currently have, let me know!

Here is an example of a function that gets a datetime.datetime object and returns a unique string for each quarter:

from datetime import datetime, timedelta

def get_quarter(d):
    return "Q%d_%d" % (math.ceil(d.month/3), d.year)

d = datetime.now()
print(d.strftime("%Y-%m-%d"), get_q(d))

d2 = d - timedelta(90)
print(d2.strftime("%Y-%m-%d"), get_q(d2))

d3 = d - timedelta(180 + 365)
print(d3.strftime("%Y-%m-%d"), get_q(d3))

And the output is:

2019-02-14 Q1_2019
2018-11-16 Q4_2018
2017-08-18 Q3_2017

if m is the month number…

import math
math.ceil(float(m) / 3)

This method works for any mapping:

month2quarter = {
        1:1,2:1,3:1,
        4:2,5:2,6:2,
        7:3,8:3,9:3,
        10:4,11:4,12:4,
    }.get

We have just generated a function int->int

month2quarter(9) # returns 3

This method is also fool-proof

month2quarter(-1) # returns None
month2quarter('July') # returns None

For those, who are looking for financial year quarter data, using pandas,

import datetime
import pandas as pd
today_date = datetime.date.today()
quarter = pd.PeriodIndex(today_date, freq='Q-MAR').strftime('Q%q')

reference: pandas period index

This is an old question but still worthy of discussion.

Here is my solution, using the excellent dateutil module.

  from dateutil import rrule,relativedelta

   year = this_date.year
   quarters = rrule.rrule(rrule.MONTHLY,
                      bymonth=(1,4,7,10),
                      bysetpos=-1,
                      dtstart=datetime.datetime(year,1,1),
                      count=8)

   first_day = quarters.before(this_date)
   last_day =  (quarters.after(this_date)
                -relativedelta.relativedelta(days=1)

So first_day is the first day of the quarter, and last_day is the last day of the quarter (calculated by finding the first day of the next quarter, minus one day).

This is very simple and works in python3:

from datetime import datetime

# Get current date-time.
now = datetime.now()

# Determine which quarter of the year is now. Returns q1, q2, q3 or q4.
quarter_of_the_year = 'q'+str((now.month-1)//3+1)

hmmm so calculations can go wrong, here is a better version (just for the sake of it)

first, second, third, fourth=1,2,3,4# you can make strings if you wish :)

quarterMap = {}
quarterMap.update(dict(zip((1,2,3),(first,)*3)))
quarterMap.update(dict(zip((4,5,6),(second,)*3)))
quarterMap.update(dict(zip((7,8,9),(third,)*3)))
quarterMap.update(dict(zip((10,11,12),(fourth,)*3)))

print quarterMap[6]

Here is a verbose, but also readable solution that will work for datetime and date instances

def get_quarter(date):
    for months, quarter in [
        ([1, 2, 3], 1),
        ([4, 5, 6], 2),
        ([7, 8, 9], 3),
        ([10, 11, 12], 4)
    ]:
        if date.month in months:
            return quarter

using dictionaries, you can pull this off by

def get_quarter(month):
    quarter_dictionary = {
        "Q1" : [1,2,3],
        "Q2" : [4,5,6],
        "Q3" : [7,8,9],
        "Q4" : [10,11,12]
    }

    for key,values in quarter_dictionary.items():
        for value in values:
            if value == month:
                return key

print(get_quarter(3))