Is there a built-in method for converting a date to a datetime in Python, for example getting the datetime for the midnight of the given date? The opposite conversion is easy: datetime has a .date() method.

Do I really have to manually call datetime(d.year, d.month, d.day)?

You can use datetime.combine(date, time); for the time, you create a datetime.time object initialized to midnight.

from datetime import date
from datetime import datetime

dt = datetime.combine(date.today(), datetime.min.time())

There are several ways, although I do believe the one you mention (and dislike) is the most readable one.

>>> t=datetime.date.today()
>>> datetime.datetime.fromordinal(t.toordinal())
datetime.datetime(2009, 12, 20, 0, 0)
>>> datetime.datetime(t.year, t.month, t.day)
datetime.datetime(2009, 12, 20, 0, 0)
>>> datetime.datetime(*t.timetuple()[:-4])
datetime.datetime(2009, 12, 20, 0, 0)

and so forth — but basically they all hinge on appropriately extracting info from the date object and ploughing it back into the suitable ctor or classfunction for datetime.

The accepted answer is correct, but I would prefer to avoid using datetime.min.time() because it’s not obvious to me exactly what it does. If it’s obvious to you, then more power to you. I also feel the same way about the timetuple method and the reliance on the ordering.

In my opinion, the most readable, explicit way of doing this without relying on the reader to be very familiar with the datetime module API is:

from datetime import date, datetime
today = date.today()
today_with_time = datetime(
    year=today.year, 
    month=today.month,
    day=today.day,
)

That’s my take on “explicit is better than implicit.”

You can use the date.timetuple() method and unpack operator *.

args = d.timetuple()[:6]
datetime.datetime(*args)

Today being 2016, I think the cleanest solution is provided by pandas Timestamp:

from datetime import date
import pandas as pd
d = date.today()
pd.Timestamp(d)

Timestamp is the pandas equivalent of datetime and is interchangable with it in most cases. Check:

from datetime import datetime
isinstance(pd.Timestamp(d), datetime)

But in case you really want a vanilla datetime, you can still do:

pd.Timestamp(d).to_datetime()

Timestamps are a lot more powerful than datetimes, amongst others when dealing with timezones. Actually, Timestamps are so powerful that it’s a pity they are so poorly documented…

One way to convert from date to datetime that hasn’t been mentioned yet:

from datetime import date, datetime
d = date.today()
datetime.strptime(d.strftime('%Y%m%d'), '%Y%m%d')

You can use easy_date to make it easy:

import date_converter
my_datetime = date_converter.date_to_datetime(my_date)

If you need something quick, datetime_object.date() gives you a date of a datetime object.

I am a newbie to Python. But this code worked for me which converts the specified input I provide to datetime. Here’s the code. Correct me if I’m wrong.

import sys
from datetime import datetime
from time import mktime, strptime

user_date = '02/15/1989'
if user_date is not None:
     user_date = datetime.strptime(user_date,"%m/%d/%Y")
else:
     user_date = datetime.now()
print user_date

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

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, 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]>

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

This is my code:

import datetime
today = datetime.date.today()
print(today)

This prints: 2008-11-22 which is exactly what I want.

But, I have a list I’m appending this to and then suddenly everything goes “wonky”. Here is the code:

import datetime
mylist = []
today = datetime.date.today()
mylist.append(today)
print(mylist)

This prints the following:

[datetime.date(2008, 11, 22)]

How can I get just a simple date like 2008-11-22?

The WHY: dates are objects

In Python, dates are objects. Therefore, when you manipulate them, you manipulate objects, not strings, not timestamps nor anything.

Any object in Python have TWO string representations:

• The regular representation that is used by “print”, can be get using the str() function. It is most of the time the most common human readable format and is used to ease display. So str(datetime.datetime(2008, 11, 22, 19, 53, 42)) gives you '2008-11-22 19:53:42'.

• The alternative representation that is used to represent the object nature (as a data). It can be get using the repr() function and is handy to know what kind of data your manipulating while you are developing or debugging. repr(datetime.datetime(2008, 11, 22, 19, 53, 42)) gives you 'datetime.datetime(2008, 11, 22, 19, 53, 42)'.

What happened is that when you have printed the date using “print”, it used str() so you could see a nice date string. But when you have printed mylist, you have printed a list of objects and Python tried to represent the set of data, using repr().

The How: what do you want to do with that?

Well, when you manipulate dates, keep using the date objects all long the way. They got thousand of useful methods and most of the Python API expect dates to be objects.

When you want to display them, just use str(). In Python, the good practice is to explicitly cast everything. So just when it’s time to print, get a string representation of your date using str(date).

One last thing. When you tried to print the dates, you printed mylist. If you want to print a date, you must print the date objects, not their container (the list).

E.G, you want to print all the date in a list :

for date in mylist :
    print str(date)

Note that in that specific case, you can even omit str() because print will use it for you. But it should not become a habit :-)

Practical case, using your code

import datetime
mylist = []
today = datetime.date.today()
mylist.append(today)
print mylist[0] # print the date object, not the container ;-)
2008-11-22

# It's better to always use str() because :

print "This is a new day : ", mylist[0] # will work
>>> This is a new day : 2008-11-22

print "This is a new day : " + mylist[0] # will crash
>>> cannot concatenate 'str' and 'datetime.date' objects

print "This is a new day : " + str(mylist[0]) 
>>> This is a new day : 2008-11-22

Advanced date formatting

Dates have a default representation, but you may want to print them in a specific format. In that case, you can get a custom string representation using the strftime() method.

strftime() expects a string pattern explaining how you want to format your date.

E.G :

print today.strftime('We are the %d, %b %Y')
>>> 'We are the 22, Nov 2008'

All the letter after a "%" represent a format for something :

• %d is the day number
• %m is the month number
• %b is the month abbreviation
• %y is the year last two digits
• %Y is the all year

etc

Have a look at the official documentation, or McCutchen’s quick reference you can’t know them all.

Since PEP3101, every object can have its own format used automatically by the method format of any string. In the case of the datetime, the format is the same used in strftime. So you can do the same as above like this:

print "We are the {:%d, %b %Y}".format(today)
>>> 'We are the 22, Nov 2008'

The advantage of this form is that you can also convert other objects at the same time.
With the introduction of Formatted string literals (since Python 3.6, 2016-12-23) this can be written as

import datetime
f"{datetime.datetime.now():%Y-%m-%d}"
>>> '2017-06-15'

Localization

Dates can automatically adapt to the local language and culture if you use them the right way, but it’s a bit complicated. Maybe for another question on SO(Stack Overflow) ;-)

import datetime
print datetime.datetime.now().strftime("%Y-%m-%d %H:%M")

Edit:

After Cees suggestion, I have started using time as well:

import time
print time.strftime("%Y-%m-%d %H:%M")

The date, datetime, and time objects all support a strftime(format) method, to create a string representing the time under the control of an explicit format string.

Here is a list of the format codes with their directive and meaning.

    %a  Locale’s abbreviated weekday name.
    %A  Locale’s full weekday name.      
    %b  Locale’s abbreviated month name.     
    %B  Locale’s full month name.
    %c  Locale’s appropriate date and time representation.   
    %d  Day of the month as a decimal number [01,31].    
    %f  Microsecond as a decimal number [0,999999], zero-padded on the left
    %H  Hour (24-hour clock) as a decimal number [00,23].    
    %I  Hour (12-hour clock) as a decimal number [01,12].    
    %j  Day of the year as a decimal number [001,366].   
    %m  Month as a decimal number [01,12].   
    %M  Minute as a decimal number [00,59].      
    %p  Locale’s equivalent of either AM or PM.
    %S  Second as a decimal number [00,61].
    %U  Week number of the year (Sunday as the first day of the week)
    %w  Weekday as a decimal number [0(Sunday),6].   
    %W  Week number of the year (Monday as the first day of the week)
    %x  Locale’s appropriate date representation.    
    %X  Locale’s appropriate time representation.    
    %y  Year without century as a decimal number [00,99].    
    %Y  Year with century as a decimal number.   
    %z  UTC offset in the form +HHMM or -HHMM.
    %Z  Time zone name (empty string if the object is naive).    
    %%  A literal '%' character.

This is what we can do with the datetime and time modules in Python

    import time
    import datetime

    print "Time in seconds since the epoch: %s" %time.time()
    print "Current date and time: ", datetime.datetime.now()
    print "Or like this: ", datetime.datetime.now().strftime("%y-%m-%d-%H-%M")


    print "Current year: ", datetime.date.today().strftime("%Y")
    print "Month of year: ", datetime.date.today().strftime("%B")
    print "Week number of the year: ", datetime.date.today().strftime("%W")
    print "Weekday of the week: ", datetime.date.today().strftime("%w")
    print "Day of year: ", datetime.date.today().strftime("%j")
    print "Day of the month : ", datetime.date.today().strftime("%d")
    print "Day of week: ", datetime.date.today().strftime("%A")

That will print out something like this:

    Time in seconds since the epoch:    1349271346.46
    Current date and time:              2012-10-03 15:35:46.461491
    Or like this:                       12-10-03-15-35
    Current year:                       2012
    Month of year:                      October
    Week number of the year:            40
    Weekday of the week:                3
    Day of year:                        277
    Day of the month :                  03
    Day of week:                        Wednesday

Use date.strftime. The formatting arguments are described in the documentation.

This one is what you wanted:

some_date.strftime('%Y-%m-%d')

This one takes Locale into account. (do this)

some_date.strftime('%c')

This is shorter:

>>> import time
>>> time.strftime("%Y-%m-%d %H:%M")
'2013-11-19 09:38'

# convert date time to regular format.

d_date = datetime.datetime.now()
reg_format_date = d_date.strftime("%Y-%m-%d %I:%M:%S %p")
print(reg_format_date)

# some other date formats.
reg_format_date = d_date.strftime("%d %B %Y %I:%M:%S %p")
print(reg_format_date)
reg_format_date = d_date.strftime("%Y-%m-%d %H:%M:%S")
print(reg_format_date)

OUTPUT

2016-10-06 01:21:34 PM
06 October 2016 01:21:34 PM
2016-10-06 13:21:34

Or even

from datetime import datetime, date

"{:%d.%m.%Y}".format(datetime.now())

Out: ‘25.12.2013

or

"{} - {:%d.%m.%Y}".format("Today", datetime.now())

Out: ‘Today – 25.12.2013’

"{:%A}".format(date.today())

Out: ‘Wednesday’

'{}__{:%Y.%m.%d__%H-%M}.log'.format(__name__, datetime.now())

Out: ‘__main____2014.06.09__16-56.log’

Simple answer –

datetime.date.today().isoformat()

With type-specific datetime string formatting (see nk9’s answer using str.format().) in a Formatted string literal (since Python 3.6, 2016-12-23):

>>> import datetime
>>> f"{datetime.datetime.now():%Y-%m-%d}"
'2017-06-15'

The date/time format directives are not documented as part of the Format String Syntax but rather in date, datetime, and time‘s strftime() documentation. The are based on the 1989 C Standard, but include some ISO 8601 directives since Python 3.6.

You need to convert the date time object to a string.

The following code worked for me:

import datetime
collection = []
dateTimeString = str(datetime.date.today())
collection.append(dateTimeString)
print collection

Let me know if you need any more help.

You can do:

mylist.append(str(today))

I hate the idea of importing too many modules for convenience. I would rather work with available module which in this case is datetime rather than calling a new module time.

>>> a = datetime.datetime(2015, 04, 01, 11, 23, 22)
>>> a.strftime('%Y-%m-%d %H:%M')
'2015-04-01 11:23'

Considering the fact you asked for something simple to do what you wanted, you could just:

import datetime
str(datetime.date.today())

For those wanting locale-based date and not including time, use:

>>> some_date.strftime('%x')
07/11/2019

You may want to append it as a string?

import datetime 
mylist = [] 
today = str(datetime.date.today())
mylist.append(today) 
print mylist

Since the print today returns what you want this means that the today object’s __str__ function returns the string you are looking for.

So you can do mylist.append(today.__str__()) as well.

You can use easy_date to make it easy:

import date_converter
my_date = date_converter.date_to_string(today, '%Y-%m-%d')

A quick disclaimer for my answer – I’ve only been learning Python for about 2 weeks, so I am by no means an expert; therefore, my explanation may not be the best and I may use incorrect terminology. Anyway, here it goes.

I noticed in your code that when you declared your variable today = datetime.date.today() you chose to name your variable with the name of a built-in function.

When your next line of code mylist.append(today) appended your list, it appended the entire string datetime.date.today(), which you had previously set as the value of your today variable, rather than just appending today().

A simple solution, albeit maybe not one most coders would use when working with the datetime module, is to change the name of your variable.

Here’s what I tried:

import datetime
mylist = []
present = datetime.date.today()
mylist.append(present)
print present

and it prints yyyy-mm-dd.

Here is how to display the date as (year/month/day) :

from datetime import datetime
now = datetime.now()

print '%s/%s/%s' % (now.year, now.month, now.day)

from datetime import date
def time-format():
  return str(date.today())
print (time-format())

this will print 6-23-2018 if that’s what you want :)

import datetime
import time

months = ["Unknown","January","Febuary","Marchh","April","May","June","July","August","September","October","November","December"]
datetimeWrite = (time.strftime("%d-%m-%Y "))
date = time.strftime("%d")
month= time.strftime("%m")
choices = {'01': 'Jan', '02':'Feb','03':'Mar','04':'Apr','05':'May','06': 'Jun','07':'Jul','08':'Aug','09':'Sep','10':'Oct','11':'Nov','12':'Dec'}
result = choices.get(month, 'default')
year = time.strftime("%Y")
Date = date+"-"+result+"-"+year
print Date

In this way you can get Date formatted like this example: 22-Jun-2017

I don’t fully understand but, can use pandas for getting times in right format:

>>> import pandas as pd
>>> pd.to_datetime('now')
Timestamp('2018-10-07 06:03:30')
>>> print(pd.to_datetime('now'))
2018-10-07 06:03:47
>>> pd.to_datetime('now').date()
datetime.date(2018, 10, 7)
>>> print(pd.to_datetime('now').date())
2018-10-07
>>> 

And:

>>> l=[]
>>> l.append(pd.to_datetime('now').date())
>>> l
[datetime.date(2018, 10, 7)]
>>> map(str,l)
<map object at 0x0000005F67CCDF98>
>>> list(map(str,l))
['2018-10-07']

But it’s storing strings but easy to convert:

>>> l=list(map(str,l))
>>> list(map(pd.to_datetime,l))
[Timestamp('2018-10-07 00:00:00')]