Difference between two dates in Python

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.

Answers:

Thank you for visiting the Q&A section on Magenaut. Please note that all the answers may not help you solve the issue immediately. So please treat them as advisements. If you found the post helpful (or not), leave a comment & I’ll get back to you as soon as possible.

Method 1

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)

Method 2

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()

Method 3

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

Method 4

You can use the third-party library dateutil, which is an extension for the built-in datetime.

Parsing dates with the parser module is very straightforward:

from dateutil import parser

date1 = parser.parse('2019-08-01')
date2 = parser.parse('2019-08-20')

diff = date2 - date1

print(diff)
print(diff.days)

Answer based on the one from this deleted duplicate

Method 5

Try this:

data=pd.read_csv('C:UsersDesktopData 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()

Method 6

I tried a couple of codes, but end up using something as simple as (in Python 3):

from datetime import datetime
df['difference_in_datetime'] = abs(df['end_datetime'] - df['start_datetime'])

If your start_datetime and end_datetime columns are in datetime64[ns] format, datetime understands it and return the difference in days + timestamp, which is in timedelta64[ns] format.

If you want to see only the difference in days, you can separate only the date portion of the start_datetime and end_datetime by using (also works for the time portion):

df['start_date'] = df['start_datetime'].dt.date
df['end_date'] = df['end_datetime'].dt.date

And then run:

df['difference_in_days'] = abs(df['end_date'] - df['start_date'])

Method 7

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


All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0

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