I'm trying to normalize a date string: '6-3-1975' to a datetime object in this format: '06/03/1975'
I have this method:
def normalizeDate(date):
formatted_date = date.replace('-', '/')
date_obj = datetime.strptime(formatted_date, '%m/%d/%Y').date()
# date = datetime.strftime(date_obj, '%m/%d/%Y')
# print(date)
return date_obj
When printing out .date() the format uses dashes, I also tried using strftime but that would convert the date_object back into a string. Is there a way to reformat the date to use slashes and still be a datetime object?
You can process the date string yourself, to give it the desired format and then convert it to datetime object:
def normalizeDate(date):
newDate = '/'.join(str.zfill(elem,2) for elem in date.split('-'))
date_obj = datetime.strptime(newDate, '%m/%d/%Y').date()
return date_obj
In order to print your date to that specific format, you will have to use strftime to the new datetime object created:
inDate = '6-3-1975'
d = normalizeDate(inDate)
print(d.strftime('%m/%d/%Y'))
print(type(d))
This is the only way to set the format of a datetime object.
Related
I have the following string 20211208_104755, representing date_time format. I want to convert it to the python datetime format using datetime.strip() method.
mydatetime = "20211208_104755"
datetime_object = datetime.strptime(mydatetime, '%y/%m/%d')
However I am getting the following error.
ValueError: time data '20211208' does not match format '%y/%m/%d'
The second argument in strptime has to match the pattern of your datetime string. You can find the patterns and their meaning on https://docs.python.org/3/library/datetime.html
In your case, you can format it as
from datetime import datetime
mydatetime = "20211208_104755"
datetime_object = datetime.strptime(mydatetime, '%Y%m%d_%H%M%S')
print(datetime_object)
>>> 2021-12-08 10:47:55
what should work is to define how the mydatetime string is composed.
example:
%Y is the year (4 digits); check here for format (section strftime() Date Format Codes)
So in your example I would assume it's like this:
mydatetime = "20211208_104755"
datetime_object = datetime.strptime(mydatetime, '%Y%m%d_%H%M%S')
print (datetime_object)
result
2021-12-08 10:47:55
and
type(datetime_object)
datetime.datetime
The below code returns as format 2021-04-30 10:30:00+12:00
time_now = datetime.now(timezone(settings.TIME_ZONE)).replace(microsecond=0)
How to get it in format 2021-04-30T10:30:00+12:00 please?
You can use astimezone(). For example:
from datetime import datetime
time_now = datetime.now().astimezone()
my_format = time_now.strftime("%Y-%m-%dT%H:%M:%S%z")
%z is empty string if object is naive. That's why you can't print it.
I using:
s = "20200113"
final = datetime.datetime.strptime(s, '%Y%m%d')
I need convert a number in date format (2020-01-13)
but when I print final:
2020-01-13 00:00:00
Tried datetime.date(s, '%Y%m%d') but It's returns a error:
an integer is required (got type str)
Is there any command to get only date without hour?
Once you have a datetime object just use strftime
import datetime
d = datetime.datetime.now() # Some datetime object.
print(d.strftime('%Y-%m-%d'))
which gives
2020-02-20
You can use strftime to convert back in the format you need :
import datetime
s = "20200113"
temp = datetime.datetime.strptime(s, '%Y%m%d')
# 2020-01-13 00:00:00
final = temp.strftime('%Y-%m-%d')
print(final)
# 2020-01-13
Use datetime.date(year, month, day). Slice your string and convert to integers to get the year, month and day. Now it is a datetime.date object, you can use it for other things. Here, however, we use .strftime to convert it back to text in your desired format.
s = "20200113"
year = int(s[:4]) # 2020
month = int(s[4:6]) # 1
day = int(s[6:8]) # 13
>>> datetime.date(year, month, day).strftime('%Y-%m-%d')
'2020-01-13'
You can also convert directly via strings.
>>> f'{s[:4]}-{s[4:6]}-{s[6:8]}'
'2020-01-13'
You can use .date() on datetime objects to 'remove' the time.
my_time_str = str(final.date())
will give you the wanted result
I am extracting Data from Mongodb using some date filter. In mongo my date is in ISO format . As i am dynamically adding date from some variable which is in timestamp format(2019-07-15 14:54:53).Getting Empty Result
curs = col1.aggregate([{'$match':{update_col: {'$gte': last_updt }}},{'$project':json_acceptable_string}])
I am expecting Rows after filtering but acual its giving empty dataset
you can use datetime.strptime to parse the original string to a datetime object, then use datetime.isoformat to get it in ISO format.
try this:
import datetime
original_date = '2019-07-15 14:54:53'
date_obj = datetime.datetime.strptime(original_date, "%Y-%m-%d %H:%M:%S")
iso_date = date_obj.isoformat()
print(iso_date)
try this
from dateutil import parser as date_parser
dt_obj = date_parser.parse('2019-07-15 14:54:53')
where dt_obj is an object of standard datetime.datetime class
You can use fromisoformat.
Try
from datetime import datetime
iso_string = '2019-07-15 14:54:53'
you_date_obj = datetime.fromisoformat(iso_string)
I have data with the date format as follows:
date_format = 190410
year = 19
month = 04
date = 10
I want to change the date format, to be like this:
date_format = 10-04-2019
How do I solve this problem?
>>> import datetime
>>> date = 190410
>>> datetime.datetime.strptime(str(date), "%y%m%d").strftime("%d-%m-%Y")
'10-04-2019'
datetime.strptime() takes a data string and a format, and turns that into datetime object, and datetime objects have a method called strftime that turns datetime objects to string with given format. You can look what %y %m %d %Y are from here.
This is what you want(Notice that you have to change your format)
import datetime
date_format = '2019-04-10'
date_time_obj = datetime.datetime.strptime(date_format, '%Y-%m-%d')
print(date_time_obj)
Here is an other example
import datetime
date_time_str = '2018-06-29 08:15:27.243860'
date_time_obj = datetime.datetime.strptime(date_time_str, '%Y-%m-%d %H:%M:%S.%f')
print('Date:', date_time_obj.date())
print('Time:', date_time_obj.time())
print('Date-time:', date_time_obj)
You can also do this
from datetime import datetime, timedelta
s = "20120213"
# you could also import date instead of datetime and use that.
date = datetime(year=int(s[0:4]), month=int(s[4:6]), day=int(s[6:8]))
print(date)
There are many ways to achieve what you want.