How to print a date in a regular format?

Question

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

This prints [datetime.date(2008, 11, 22)]. How can I get just a simple date like 2008-11-22?

Answer 1

The WHY: dates are objects

In Python, dates are objects. Therefore, when you manipulate them, you manipulate objects, not strings or timestamps.

Any object in Python has 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 (2 digits, prefixed with leading zero's if necessary)
• %m is the month number (2 digits, prefixed with leading zero's if necessary)
• %b is the month abbreviation (3 letters)
• %B is the month name in full (letters)
• %y is the year number abbreviated (last 2 digits)
• %Y is the year number full (4 digits)

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) ;-)

Answer 2

You can also use the built-in format() function to format a date/datetime. For example:

import datetime
format(datetime.date.today(), "%Y-%m-%d")    # '2023-12-05'

This is especially useful if you have a use case where you need a callable to format a list/sequence of date/datetimes. For example, for the case in the OP, we can map the format function to mylist as follows:

mylist = [datetime.date(2008, 11, 22), datetime.datetime(2023, 12, 5, 11, 30, 5)]

list(map(format, mylist, ['%Y-%m-%d']*len(mylist)))   # ['2008-11-22', '2023-12-05']

or using the built-in itertools module, it can be written a little more efficiently as follows:

from itertools import repeat
list(map(format, mylist, repeat('%Y-%m-%d')))         # ['2008-11-22', '2023-12-05']

Another use case where it comes in useful is when we want to convert a pandas column of datetimes and format them into strings. The pandas built-in dt.strftime is really slow but applying format() is much faster.

import pandas as pd
pd.Series(mylist).apply(format, args=["%Y-%m-%d"])

Answer 3

maybe the shortest solution, which exactly matches your situation, would be:

mylist.append(str(AnyDate)[:10])

or even shorter, e.g.:

f'{AnyDate}'[:10]

PS: it doesn't need to be today.

Answer 4

For pandas.Timestamps, strftime() can be used e.g.:

utc_now = datetime.now()

For isoformat:

utc_now.isoformat()

For any format e.g.:

utc_now.strftime("%m/%d/%Y, %H:%M:%S")

Answer 5

In Python you can format a datetime using the strftime() method from the date, time and datetime classes in the datetime module.

In your specific case, you are using the date class from datetime. You can use the following snippet to format the today variable into a string with the format yyyy-MM-dd:

import datetime

today = datetime.date.today()
print("formatted datetime: %s" % today.strftime("%Y-%m-%d"))

In the following a more complete example:

import datetime
today = datetime.date.today()

# datetime in d/m/Y H:M:S format
date_time = today.strftime("%d/%m/%Y, %H:%M:%S")
print("datetime: %s" % date_time)

# datetime in Y-m-d H:M:S format
date_time = today.strftime("%Y-%m-%d, %H:%M:%S")
print("datetime: %s" % date_time)

# format date
date = today.strftime("%d/%m/%Y")
print("date: %s" % time)

# format time
time = today.strftime("%H:%M:%S")
print("time: %s" % time)

# day
day = today.strftime("%d")
print("day: %s" % day)

# month
month = today.strftime("%m")
print("month: %s" % month)

# year
year = today.strftime("%Y")
print("year: %s" % year)

More directives:

Sources:

• Format DateTime in Python
• strftime

Answer 6

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

Answer 7

You need to convert the datetime object to a str.

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.

Answer 8

You may want to append it as a string?

import datetime

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

print(mylist)

Answer 9

from datetime import date

def today_in_str_format():
    return str(date.today())

print (today_in_str_format())

This will print 2018-06-23 if that's what you want :)

Answer 10

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")

Answer 11

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

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

Answer 12

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')]

Answer 13

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

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

Answer 14

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'

Answer 15

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

Answer 16

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.

Answer 17

# 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

Answer 18

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)

Answer 19

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.

Answer 20

You can use easy_date to make it easy:

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

Answer 21

Simple answer -

datetime.date.today().isoformat()

Answer 22

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.

Answer 23

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'

Answer 24

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')

Answer 25

This is shorter:

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

Answer 26

You can do:

mylist.append(str(today))