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
?
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'
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')
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')]
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'
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
.
This is shorter:
>>> import time
>>> time.strftime("%Y-%m-%d %H:%M")
'2013-11-19 09:38'
For those wanting locale-based date and not including time, use:
>>> some_date.strftime('%x')
07/11/2019
Simple answer -
datetime.date.today().isoformat()
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')
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.
# 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
Simple answer -
datetime.date.today().isoformat()
You can do:
mylist.append(str(today))
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
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.
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")
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.
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 :)
Considering the fact you asked for something simple to do what you wanted, you could just:
import datetime
str(datetime.date.today())
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')]
# 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
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')
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 may want to append it as a string?
import datetime
mylist = []
today = str(datetime.date.today())
mylist.append(today)
print mylist
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')
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'
You may want to append it as a string?
import datetime
mylist = []
today = str(datetime.date.today())
mylist.append(today)
print mylist
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)
Considering the fact you asked for something simple to do what you wanted, you could just:
import datetime
str(datetime.date.today())
You can do:
mylist.append(str(today))
You may want to append it as a string?
import datetime
mylist = []
today = str(datetime.date.today())
mylist.append(today)
print mylist
For those wanting locale-based date and not including time, use:
>>> some_date.strftime('%x')
07/11/2019
You can do:
mylist.append(str(today))
This is shorter:
>>> import time
>>> time.strftime("%Y-%m-%d %H:%M")
'2013-11-19 09:38'
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
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 :)
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'
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))
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)
You may want to append it as a string?
import datetime
mylist = []
today = str(datetime.date.today())
mylist.append(today)
print mylist
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")
You can use easy_date to make it easy:
import date_converter
my_date = date_converter.date_to_string(today, '%Y-%m-%d')
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.
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')
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
Source: Stackoverflow.com