Is there a direct way to ignore parts of a python datetime object?

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Chapters
00:00 Is There A Direct Way To Ignore Parts Of A Python Datetime Object?
00:37 Answer 1 Score 9
00:52 Answer 2 Score 2
01:09 Answer 3 Score 4
01:31 Accepted Answer Score 6
01:38 Thank you

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Full question
https://stackoverflow.com/questions/3915...

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Content licensed under CC BY-SA
https://meta.stackexchange.com/help/lice...

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Tags
#python #datetime

#avk47

ANSWER 1

Score 9

(a.month, a.day, a.hour, a.minute, a.second == 
 b.month, b.day, b.hour, b.minute, b.second)

A less explicit method is to compare the corresponding elements in the time tuples:

a.timetuple()[1:6] == b.timetuple()[1:6]

ACCEPTED ANSWER

Score 6

Try:

a.replace(year=1,microsecond=0) == b.replace(year=1,microsecond=0)

ANSWER 3

Score 4

You can also consider comparing formatted date strings consisting of the fields you wish to include in the comparison. This allows you to be somewhat explicit while using shortened versions of the fields (as opposed to accessing a.year, a.month, etc.).

from datetime import datetime

date_string = '%m %d %H %M %S'
a = datetime(2015, 7, 4, 1, 1, 1)
b = datetime(2016, 7, 4, 1, 1, 1)
print(a.strftime(date_string) == b.strftime(date_string))  # True

ANSWER 4

Score 2

def cmp(a,b):
 return (a > b) - (a < b)

d1=(2015,7,4,1,1,1)
d2=(2016,7,4,1,1,1)

cmp(list(d1)[1:],list(d2)[1:])

Returns 0 - they are the same, i.e. 0 differences

d1=(2015,7,4,1,1,1)
d2=(2015,2,4,1,1,1)

cmp(list(d1)[1:], list(d2)[1:])

returns -1, there is a difference.