Python: Difference of 2 datetimes in months
Question:
Possible Duplicate:
Best way to find the months between two dates (in python)
I would like to know how I can have the exact number of months for this difference:
date1 = datetime.strptime(str('2011-08-15 12:00:00'), '%Y-%m-%d %H:%M:%S')
date2 = datetime.strptime(str('2012-02-15'), '%Y-%m-%d')
date2-date1 results in
datetime.timedelta(183, 43200)
I would like to know the exact number of months, in this case it should return 5 and not 6 (because of the hour)
Answers:
The advantage of doing it this way is that there are few module dependencies and no looping — the months can be found by straight calculation.
import datetime as dt
def months_between(date1,date2):
if date1>date2:
date1,date2=date2,date1
m1=date1.year*12+date1.month
m2=date2.year*12+date2.month
months=m2-m1
if date1.day>date2.day:
months-=1
elif date1.day==date2.day:
seconds1=date1.hour*3600+date1.minute+date1.second
seconds2=date2.hour*3600+date2.minute+date2.second
if seconds1>seconds2:
months-=1
return months
date1 = dt.datetime.strptime('2011-08-15 12:00:00', '%Y-%m-%d %H:%M:%S')
date2 = dt.datetime.strptime('2012-02-15', '%Y-%m-%d')
print(months_between(date1,date2))
# 5
date1 = dt.datetime.strptime('2011-08-15 12:00:00', '%Y-%m-%d %H:%M:%S')
date2 = dt.datetime.strptime('2012-02-15 11:59:00', '%Y-%m-%d %X')
print(months_between(date1,date2))
# 5
date2 = dt.datetime.strptime('2012-02-15 12:00:00', '%Y-%m-%d %X')
print(months_between(date1,date2))
# 6
date2 = dt.datetime.strptime('2012-02-15 12:00:01', '%Y-%m-%d %X')
print(months_between(date1,date2))
# 6
You could use python-dateutil.
In [4]: from datetime import datetime
In [5]: date1 = datetime.strptime(str('2011-08-15 12:00:00'), '%Y-%m-%d %H:%M:%S')
In [6]: date2 = datetime.strptime(str('2012-02-15'), '%Y-%m-%d')
In [7]: from dateutil import relativedelta
In [8]: r = relativedelta.relativedelta(date1, date2)
In [9]: r
Out[9]: relativedelta(months=-5, days=-30, hours=-12)
Using calendar
module to find out how many days each month has, you can simply count the months.
from calendar import monthrange
from datetime import datetime, timedelta
def monthdelta(d1, d2):
delta = 0
while True:
mdays = monthrange(d1.year, d1.month)[1]
d1 += timedelta(days=mdays)
if d1 <= d2:
delta += 1
else:
break
return delta
Only you know the requirements you must meet, but the fact that there are 183 days and 43200 SI seconds between these two dates highlights an inherent subjectivity in determining how many months that “really” is.
Is a month 30 days, or (365 / 12) days, or ((365 * 4 + 1) / 48) days, or …?
Is a day always 86400 seconds, or do you count historical leap seconds, or do you predict leap seconds for future dates?
These decisions affect the answer the algorithm you appear to desire will give you for certain input dates that are close to these boundaries.
In my opinion, it is more intuitive to consider months as atomic units of time for this purpose and use this formula: (date2.year - date1.year) * 12 + (date2.month - date1.month)
Possible Duplicate:
Best way to find the months between two dates (in python)
I would like to know how I can have the exact number of months for this difference:
date1 = datetime.strptime(str('2011-08-15 12:00:00'), '%Y-%m-%d %H:%M:%S')
date2 = datetime.strptime(str('2012-02-15'), '%Y-%m-%d')
date2-date1 results in
datetime.timedelta(183, 43200)
I would like to know the exact number of months, in this case it should return 5 and not 6 (because of the hour)
The advantage of doing it this way is that there are few module dependencies and no looping — the months can be found by straight calculation.
import datetime as dt
def months_between(date1,date2):
if date1>date2:
date1,date2=date2,date1
m1=date1.year*12+date1.month
m2=date2.year*12+date2.month
months=m2-m1
if date1.day>date2.day:
months-=1
elif date1.day==date2.day:
seconds1=date1.hour*3600+date1.minute+date1.second
seconds2=date2.hour*3600+date2.minute+date2.second
if seconds1>seconds2:
months-=1
return months
date1 = dt.datetime.strptime('2011-08-15 12:00:00', '%Y-%m-%d %H:%M:%S')
date2 = dt.datetime.strptime('2012-02-15', '%Y-%m-%d')
print(months_between(date1,date2))
# 5
date1 = dt.datetime.strptime('2011-08-15 12:00:00', '%Y-%m-%d %H:%M:%S')
date2 = dt.datetime.strptime('2012-02-15 11:59:00', '%Y-%m-%d %X')
print(months_between(date1,date2))
# 5
date2 = dt.datetime.strptime('2012-02-15 12:00:00', '%Y-%m-%d %X')
print(months_between(date1,date2))
# 6
date2 = dt.datetime.strptime('2012-02-15 12:00:01', '%Y-%m-%d %X')
print(months_between(date1,date2))
# 6
You could use python-dateutil.
In [4]: from datetime import datetime
In [5]: date1 = datetime.strptime(str('2011-08-15 12:00:00'), '%Y-%m-%d %H:%M:%S')
In [6]: date2 = datetime.strptime(str('2012-02-15'), '%Y-%m-%d')
In [7]: from dateutil import relativedelta
In [8]: r = relativedelta.relativedelta(date1, date2)
In [9]: r
Out[9]: relativedelta(months=-5, days=-30, hours=-12)
Using calendar
module to find out how many days each month has, you can simply count the months.
from calendar import monthrange
from datetime import datetime, timedelta
def monthdelta(d1, d2):
delta = 0
while True:
mdays = monthrange(d1.year, d1.month)[1]
d1 += timedelta(days=mdays)
if d1 <= d2:
delta += 1
else:
break
return delta
Only you know the requirements you must meet, but the fact that there are 183 days and 43200 SI seconds between these two dates highlights an inherent subjectivity in determining how many months that “really” is.
Is a month 30 days, or (365 / 12) days, or ((365 * 4 + 1) / 48) days, or …?
Is a day always 86400 seconds, or do you count historical leap seconds, or do you predict leap seconds for future dates?
These decisions affect the answer the algorithm you appear to desire will give you for certain input dates that are close to these boundaries.
In my opinion, it is more intuitive to consider months as atomic units of time for this purpose and use this formula: (date2.year - date1.year) * 12 + (date2.month - date1.month)