I have situation where I should ignore the timestamp in a date string. I have tried the below command but with no luck.
"start" variable used below is in AbsTime (Ex: 01MAY2017 11:45) and not a string.
start_date = datetime.datetime.strptime(start, '%d%^b%Y').date()
print start_date
My output should be :
01MAY2017 or 01MAY2017 00:00
Could any one help me.
4 Answers
Your directives are slightly off, and you need to capture all the contents (irrespective of what you want to retain).
start_date = datetime.datetime.strptime(start, '%d%b%Y %H:%M').date()
print start_date.strftime('%d%b%Y')
# '01May2017'
Update - Adding complete code below:
import datetime
start = '01MAY2017 11:45'
start_date = datetime.datetime.strptime(start, '%d%b%Y %H:%M').date()
print start_date.strftime('%d%b%Y')
# 01May2017
To convert the string back to a datetime we can use strptime()
Example using strptime()
datetime.datetime.strptime('10Apr2017 00:00', '%d%b%Y %H:%M')
In [17]: datetime.datetime.strptime('10Apr2017 00:00', '%d%b%Y %H:%M')
Out[17]: datetime.datetime(2017, 4, 10, 0, 0)
Use strftime to form your datetime object
Example using now() returns an datetime object that we form to a string
datetime.datetime.now().strftime('%d%b%Y')
In [14]: datetime.datetime.now().strftime('%d%b%Y')
Out[14]: '10Apr2017'
try this
import datetime
start = '01MAY2017 11:45'
start_date = datetime.datetime.strptime(start, '%d%b%Y %H:%M')
print start_date.strftime('%Y-%m-%d')
Actually My Best Option to use the very same datetime object to subtract hours , if you do not wish to use strings and use the datetime Object itself.
Here's an Example:
In [1]: date.strftime('%c')
Out[2]: 'Wed Jul 10 00:00:00 2019'
In [3]: date = datetime.datetime.utcnow()
In [4]: date.strftime('%c')
Out[5]: 'Wed Jul 10 19:06:22 2019'
In [6]: date = date - datetime.timedelta(hours = date.hour, minutes = date.minute, seconds = date.second) #Removing hours, mins,secs
In [7]: date.strftime('%c') #use
Out[8]: 'Wed Jul 10 00:00:00 2019'
