How to subtract a day from a date

Question

I have a Python datetime datetime object  What is the best way to subtract one day

User · Answer

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 now
yesterday 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]>

User · Answer

You can use a timedelta object  from datetime import datetime  timedelta      d   datetime today   - timedelta days days to subtract

User · Answer

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    gt 2015-03-01 2015-02-28

User · Answer

You can use a timedelta object  from datetime import datetime  timedelta      d   datetime today   - timedelta days days to subtract

User · Answer

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

User · Answer

You can use a timedelta object  from datetime import datetime  timedelta      d   datetime today   - timedelta days days to subtract

User · Answer

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

User · Answer

Subtract datetime timedelta days 1

User · Answer

Just to Elaborate an alternate method and a Use case for which it is helpful    Subtract 1 day from current datetime     from datetime import datetime  timedelta print datetime now     timedelta days -1     Here  I am adding a negative timedelta     Useful in the Case  If you want to add 5 days and subtract 5 hours from current datetime  i e  What is the Datetime 5 days from now but 5 hours less      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

User · Answer

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 now
yesterday 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]>

User · Answer

Subtract datetime timedelta days 1

User · Answer

You can use a timedelta object  from datetime import datetime  timedelta      d   datetime today   - timedelta days days to subtract

User · Answer

Subtract datetime timedelta days 1

User · Answer

Just to Elaborate an alternate method and a Use case for which it is helpful    Subtract 1 day from current datetime     from datetime import datetime  timedelta print datetime now     timedelta days -1     Here  I am adding a negative timedelta     Useful in the Case  If you want to add 5 days and subtract 5 hours from current datetime  i e  What is the Datetime 5 days from now but 5 hours less      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

User · Answer

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    gt 2015-03-01 2015-02-28

User · Answer

Subtract datetime timedelta days 1

[python] How to subtract a day from a date?

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