I have a Python datetime.datetime object. What is the best way to subtract one day?
If your Python datetime object is timezone-aware than you should be careful to avoid errors around DST transitions (or changes in UTC offset for other reasons):
from datetime import datetime, timedelta
from tzlocal import get_localzone # pip install tzlocal
DAY = timedelta(1)
local_tz = get_localzone() # get local timezone
now = datetime.now(local_tz) # get timezone-aware datetime object
day_ago = local_tz.normalize(now - DAY) # exactly 24 hours ago, time may differ
naive = now.replace(tzinfo=None) - DAY # same time
yesterday = local_tz.localize(naive, is_dst=None) # but elapsed hours may differ
In general, day_ago and yesterday may differ if UTC offset for the local timezone has changed in the last day.
For example, daylight saving time/summer time ends on Sun 2-Nov-2014 at 02:00:00 A.M. in America/Los_Angeles timezone therefore if:
import pytz # pip install pytz
local_tz = pytz.timezone('America/Los_Angeles')
now = local_tz.localize(datetime(2014, 11, 2, 10), is_dst=None)
# 2014-11-02 10:00:00 PST-0800
then day_ago and yesterday differ:
day_ago is exactly 24 hours ago (relative to now) but at 11 am, not at 10 am as nowyesterday is yesterday at 10 am but it is 25 hours ago (relative to now), not 24 hours.pendulum module handles it automatically:
>>> import pendulum # $ pip install pendulum
>>> now = pendulum.create(2014, 11, 2, 10, tz='America/Los_Angeles')
>>> day_ago = now.subtract(hours=24) # exactly 24 hours ago
>>> yesterday = now.subtract(days=1) # yesterday at 10 am but it is 25 hours ago
>>> (now - day_ago).in_hours()
24
>>> (now - yesterday).in_hours()
25
>>> now
<Pendulum [2014-11-02T10:00:00-08:00]>
>>> day_ago
<Pendulum [2014-11-01T11:00:00-07:00]>
>>> yesterday
<Pendulum [2014-11-01T10:00:00-07:00]>
Just to Elaborate an alternate method and a Use case for which it is helpful:
from datetime import datetime, timedelta print datetime.now() + timedelta(days=-1) # Here, I am adding a negative timedelta
from datetime import datetime, timedelta print datetime.now() + timedelta(days=5, hours=-5)
It can similarly be used with other parameters e.g. seconds, weeks etc
Also just another nice function i like to use when i want to compute i.e. first/last day of the last month or other relative timedeltas etc. ...
The relativedelta function from dateutil function (a powerful extension to the datetime lib)
import datetime as dt
from dateutil.relativedelta import relativedelta
#get first and last day of this and last month)
today = dt.date.today()
first_day_this_month = dt.date(day=1, month=today.month, year=today.year)
last_day_last_month = first_day_this_month - relativedelta(days=1)
print (first_day_this_month, last_day_last_month)
>2015-03-01 2015-02-28
Subtract datetime.timedelta(days=1)
Genial arrow module exists
import arrow
utc = arrow.utcnow()
utc_yesterday = utc.shift(days=-1)
print(utc, '\n', utc_yesterday)
output:
2017-04-06T11:17:34.431397+00:00
2017-04-05T11:17:34.431397+00:00
Subtract datetime.timedelta(days=1)
Also just another nice function i like to use when i want to compute i.e. first/last day of the last month or other relative timedeltas etc. ...
The relativedelta function from dateutil function (a powerful extension to the datetime lib)
import datetime as dt
from dateutil.relativedelta import relativedelta
#get first and last day of this and last month)
today = dt.date.today()
first_day_this_month = dt.date(day=1, month=today.month, year=today.year)
last_day_last_month = first_day_this_month - relativedelta(days=1)
print (first_day_this_month, last_day_last_month)
>2015-03-01 2015-02-28
Genial arrow module exists
import arrow
utc = arrow.utcnow()
utc_yesterday = utc.shift(days=-1)
print(utc, '\n', utc_yesterday)
output:
2017-04-06T11:17:34.431397+00:00
2017-04-05T11:17:34.431397+00:00
Just to Elaborate an alternate method and a Use case for which it is helpful:
from datetime import datetime, timedelta print datetime.now() + timedelta(days=-1) # Here, I am adding a negative timedelta
from datetime import datetime, timedelta print datetime.now() + timedelta(days=5, hours=-5)
It can similarly be used with other parameters e.g. seconds, weeks etc
If your Python datetime object is timezone-aware than you should be careful to avoid errors around DST transitions (or changes in UTC offset for other reasons):
from datetime import datetime, timedelta
from tzlocal import get_localzone # pip install tzlocal
DAY = timedelta(1)
local_tz = get_localzone() # get local timezone
now = datetime.now(local_tz) # get timezone-aware datetime object
day_ago = local_tz.normalize(now - DAY) # exactly 24 hours ago, time may differ
naive = now.replace(tzinfo=None) - DAY # same time
yesterday = local_tz.localize(naive, is_dst=None) # but elapsed hours may differ
In general, day_ago and yesterday may differ if UTC offset for the local timezone has changed in the last day.
For example, daylight saving time/summer time ends on Sun 2-Nov-2014 at 02:00:00 A.M. in America/Los_Angeles timezone therefore if:
import pytz # pip install pytz
local_tz = pytz.timezone('America/Los_Angeles')
now = local_tz.localize(datetime(2014, 11, 2, 10), is_dst=None)
# 2014-11-02 10:00:00 PST-0800
then day_ago and yesterday differ:
day_ago is exactly 24 hours ago (relative to now) but at 11 am, not at 10 am as nowyesterday is yesterday at 10 am but it is 25 hours ago (relative to now), not 24 hours.pendulum module handles it automatically:
>>> import pendulum # $ pip install pendulum
>>> now = pendulum.create(2014, 11, 2, 10, tz='America/Los_Angeles')
>>> day_ago = now.subtract(hours=24) # exactly 24 hours ago
>>> yesterday = now.subtract(days=1) # yesterday at 10 am but it is 25 hours ago
>>> (now - day_ago).in_hours()
24
>>> (now - yesterday).in_hours()
25
>>> now
<Pendulum [2014-11-02T10:00:00-08:00]>
>>> day_ago
<Pendulum [2014-11-01T11:00:00-07:00]>
>>> yesterday
<Pendulum [2014-11-01T10:00:00-07:00]>
Subtract datetime.timedelta(days=1)
Subtract datetime.timedelta(days=1)
Source: Stackoverflow.com