How to subtract a day from a date?

You can use a timedelta object:

from datetime import datetime, timedelta    d = datetime.today() - timedelta(days=days_to_subtract)

Subtract datetime.timedelta(days=1)

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, timedeltafrom tzlocal import get_localzone # pip install tzlocalDAY = timedelta(1)local_tz = get_localzone()   # get local timezonenow = datetime.now(local_tz) # get timezone-aware datetime objectday_ago = local_tz.normalize(now - DAY) # exactly 24 hours ago, time may differnaive = now.replace(tzinfo=None) - DAY # same timeyesterday = 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 pytzlocal_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]>