I have a Python datetime.datetime
object. What is the best way to subtract one day?
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2related: Find if 24 hrs have passed between datetimes - Python– jfsCommented Feb 2, 2015 at 12:52
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This thread is directly useful to convert a day of the year to a date.– HariCommented Nov 9, 2020 at 15:58
8 Answers
You can use a timedelta
object:
from datetime import datetime, timedelta
d = datetime.today() - timedelta(days=days_to_subtract)
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43if you don't ignore timezones then the answer is more complex.– jfsCommented Aug 21, 2014 at 13:39
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2Also, how do you relate it with a specific date. See my question: stackoverflow.com/questions/43092508/…– WJACommented Mar 29, 2017 at 12:10
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2It works with other units as well, I've used it with
timedelta(minutes=12)
for example.– NagevCommented Feb 16, 2018 at 17:29 -
the documentation says that will return "A duration expressing the difference between two date, time, or datetime instances...". how do you get the actual date of, say, 5 days ago or 5 days from now?– oldboyCommented Jul 20, 2018 at 3:34
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2@jfs I think this answer is fine in regard to time zones. If you subtract a day, that might actually subtract 23 or 25 hours, but the time component will remain the same. That's what I would expect as normal behavior. Daylight saving time (DST) makes so that 1 day is not always 24 hours.– at54321Commented Oct 13, 2021 at 7:51
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 tonow
) but at 11 am, not at 10 am asnow
yesterday
is yesterday at 10 am but it is 25 hours ago (relative tonow
), 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:
- 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
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
You can also use pandas pd.Timedelta('1D')
(it's extremely flexible and you can even pass something like pd.Timedelta('1d 5h 9s')
for 1 day, 5 hours and 9 seconds).
A convenient thing about pandas is that its datetime objects are built on datetime.datetime
, so any operation involving Python's datetime
objects work fine on pandas datetime objects and vice versa.
import pandas as pd
import numpy as np
from datetime import datetime, date, timedelta
datetime.now() - pd.Timedelta('1d') # datetime.datetime(2023, 2, 21, 15, 35, 23, 603832)
pd.Timestamp('now') - timedelta(days=1) # Timestamp('2023-02-21 15:35:23.741866')
pd.Timestamp('now') - pd.Timedelta('1d') # Timestamp('2023-02-21 15:35:23.882746')
pd.Timestamp('now') - np.timedelta64(1, 'D') # Timestamp('2023-02-21 15:35:24.032356')
date(2022, 2, 22) - pd.Timedelta('10d') # datetime.date(2022, 2, 12)
The advantage of pandas is that you can perform vectorized operations (even if the dtype is object). You can use either of pd.Timedelta
/datetime.timedelta
/np.timedelta64
.
pd.Series([datetime(2023,2,22), datetime(2023,2,21), datetime(2023,2,20)]) - pd.Timedelta('1d')
pd.Series([date(2023,2,22), date(2023,2,21), date(2023,2,20)]) - timedelta(days=1)
pd.Series([date(2023,2,22), date(2023,2,21), date(2023,2,20)]) - np.timedelta64(1, 'D')
class myDate:
def __init__(self):
self.day = 0
self.month = 0
self.year = 0
## for checking valid days month and year
while (True):
d = int(input("Enter The day :- "))
if (d > 31):
print("Plz 1 To 30 value Enter ........")
else:
self.day = d
break
while (True):
m = int(input("Enter The Month :- "))
if (m > 13):
print("Plz 1 To 12 value Enter ........")
else:
self.month = m
break
while (True):
y = int(input("Enter The Year :- "))
if (y > 9999 and y < 0000):
print("Plz 0000 To 9999 value Enter ........")
else:
self.year = y
break
## method for aday ands cnttract days
def adayDays(self, n):
## aday days to date day
nd = self.day + n
print(nd)
## check days subtract from date
if nd == 0: ## check if days are 7 subtracted from 7 then,........
if(self.year % 4 == 0):
if(self.month == 3):
self.day = 29
self.month -= 1
self.year = self. year
else:
if(self.month == 3):
self.day = 28
self.month -= 1
self.year = self. year
if (self.month == 5) or (self.month == 7) or (self.month == 8) or (self.month == 10) or (self.month == 12):
self.day = 30
self.month -= 1
self.year = self. year
elif (self.month == 2) or (self.month == 4) or (self.month == 6) or (self.month == 9) or (self.month == 11):
self.day = 31
self.month -= 1
self.year = self. year
elif(self.month == 1):
self.month = 12
self.year -= 1
## nd == 0 if condition over
## after subtract days to day io goes into negative then
elif nd < 0 :
n = abs(n)## return positive if no is negative
for i in range (n,0,-1): ##
if self.day == 0:
if self.month == 1:
self.day = 30
self.month = 12
self.year -= 1
else:
self.month -= 1
if(self.month == 1) or (self.month == 3)or (self.month == 5) or (self.month == 7) or (self.month == 8) or (self.month == 10) or (self.month ==12):
self.day = 30
elif(self.month == 4)or (self.month == 6) or (self.month == 9) or (self.month == 11):
self.day = 29
elif(self.month == 2):
if(self.year % 4 == 0):
self.day == 28
else:
self.day == 27
else:
self.day -= 1
## enf of elif negative days
## adaying days to DATE
else:
cnt = 0
while (True):
if self.month == 2: # check leap year
if(self.year % 4 == 0):
if(nd > 29):
cnt = nd - 29
nd = cnt
self.month += 1
else:
self.day = nd
break
## if not leap year then
else:
if(nd > 28):
cnt = nd - 28
nd = cnt
self.month += 1
else:
self.day = nd
break
## checking month other than february month
elif(self.month == 1) or (self.month == 3) or (self.month == 5) or (self.month == 7) or (self.month == 8) or (self.month == 10) or (self.month == 12):
if(nd > 31):
cnt = nd - 31
nd = cnt
if(self.month == 12):
self.month = 1
self.year += 1
else:
self.month += 1
else:
self.day = nd
break
elif(self.month == 4) or (self.month == 6) or (self.month == 9) or (self.month == 11):
if(nd > 30):
cnt = nd - 30
nd = cnt
self.month += 1
else:
self.day = nd
break
## end of month condition
## end of while loop
## end of else condition for adaying days
def formatDate(self,frmt):
if(frmt == 1):
ff=str(self.day)+"-"+str(self.month)+"-"+str(self.year)
elif(frmt == 2):
ff=str(self.month)+"-"+str(self.day)+"-"+str(self.year)
elif(frmt == 3):
ff =str(self.year),"-",str(self.month),"-",str(self.day)
elif(frmt == 0):
print("Thanky You.....................")
else:
print("Enter Correct Choice.......")
print(ff)
dt = myDate()
nday = int(input("Enter No. For Aday or SUBTRACT Days :: "))
dt.adayDays(nday)
print("1 : day-month-year")
print("2 : month-day-year")
print("3 : year-month-day")
print("0 : EXIT")
frmt = int (input("Enter Your Choice :: "))
dt.formatDate(frmt)
enter code here
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3It is better to use the standard library when possible, cause it is usually maintained by a lot of people, has little errors and performs faster Commented May 5, 2022 at 12:50
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DateTime lib do the work in here :) no need for special logic Commented Sep 8, 2022 at 9:17