I get date in format(YYYY-MM-DD)
Then I want to add timedelta
mydate + timedelta(days=1)
and I get error
coercing to Unicode: need string or buffer, datetime.timedelta found
mydate is a string. Try doing this:
from datetime import datetime
parsed_date = datetime.strptime(mydate, "%Y-%m-%d")
new_date = parsed_date + timedelta(days=1)
Data sent by the client will be sent as a Unicode and you have to parse it on server side
datetime.strptime(date, '%Y-%m-%d')
If it's part of a form, the data should be reformatted automatically when cleaned (though you might need to configure the field to accept the format you expect).
You have to convert your string date to python datetime:
>>> from datetime import datetime, timedelta
>>> dt_str = "2013/10/11"
>>> dt = datetime.strptime(dt_str, "%Y/%m/%d")
>>> new_dt = dt + timedelta(days=1)
>>> print new_dt
datetime.datetime(2013, 10, 12, 0, 0)
If you now want to get the date as string:
>>> print new_dt.strftime("%Y/%m/%d")
'2013/10/12'
Related
How can I convert string 2021-09-30_1 to datetime 2021/09/30 00:00, which means that from the last string we have to substract one to get the hour.
I tried datetime.strptime(date, '%Y %d %Y %I')
datetime.strptime if to define the timestamp from a string, the format should match the provided one. datetime.strftime (note the f) is to generate a string from a datetime object.
You can use:
datetime.strptime(date, '%Y-%m-%d_%H').strftime('%Y/%m/%d %H:%M')
output: '2021/09/30 01:00'
in case the _x defines a delta:
from datetime import datetime, timedelta
d, h = date.split('_')
d = datetime.strptime(d, '%Y-%m-%d')
h = timedelta(hours=int(h))
(d-h).strftime('%Y/%m/%d %H:%M')
output: '2021/09/29 23:00'
Considering the _1 is hour and appears in al of your data (The hour part takes value between [1, 24]), your format was wrong.
For reading the date from string you'll need format it correctly:
from datetime import datetime, timedelta
date = "2021-09-30_1"
date_part, hour_part = date.split("_")
date_object = datetime.strptime(date_part, '%Y-%m-%d') + timedelta(hours=int(hour_part) - 1)
Now you have the date object. And you can display it as:
print(date_object.strftime('%Y/%m/%d %H:%M'))
from datetime import datetime
raw_date = "2021-09-30_1"
date = raw_date.split("_")[0]
parsed_date = datetime.strptime(date, '%Y-%m-%d')
formated_date = parsed_date.strftime('%Y/%m/%d %H:%M')
strptime is used for parsing string and strftime for formating.
Also for date representation you should provide format codes for hours and minutes as in:
formated_date = parsed_date.strftime('%Y/%m/%d %H:%M')
How can one make 2020/09/06 15:59:04 out of 06-09-202015u59m04s.
This is my code:
my_time = '06-09-202014u59m04s'
date_object = datetime.datetime.strptime(my_time, '%d-%m-%YT%H:%M:%S')
print(date_object)
This is the error I receive:
ValueError: time data '06-09-202014u59m04s' does not match format '%d-%m-%YT%H:%M:%S'
>>> from datetime import datetime
>>> my_time = '06-09-202014u59m04s'
>>> dt_obj = datetime.strptime(my_time,'%d-%m-%Y%Hu%Mm%Ss')
Now you need to do some format changes to get the answer as the datetime object always prints itself with : so you can do any one of the following:
Either get a new format using strftime:
>>> dt_obj.strftime('%Y/%m/%d %H:%M:%S')
'2020/09/06 14:59:04'
Or you can simply use .replace() by converting datetime object to str:
>>> str(dt_obj).replace('-','/')
'2020/09/06 14:59:04'
As your error says what you give does not match format - %d-%m-%YT%H:%M:%S - means you are expecting after year: letter T hour:minutes:seconds when in example show it is houruminutesmsecondss without T, so you should do:
import datetime
my_time = '06-09-202014u59m04s'
date_object = datetime.datetime.strptime(my_time, '%d-%m-%Y%Hu%Mm%Ss')
print(date_object)
Output:
2020-09-06 14:59:04
You need to always make sure that your desired date format should match up with your required format.
from datetime import datetime
date_object = datetime.strptime("06-09-202015u59m04s", '%d-%m-%Y%Hu%Mm%Ss')
print(date_object.strftime('%Y/%m/%d %H:%M:%S'))
Output
2020/09/06 15:59:04
I have an API that return this date string
strdate = '2019-10-07T06:09:28.984Z'
How do I convert it to a datetime object?
dt = datetime.datetime.strptime(strdate, '%Y-%m-%d')
If I use strptime like above i get "ValueError: unconverted data remains: T06:09:54.346Z"
What do I do if there is "T" and "Z" included in the string date? I want the data in local time. But is the string really timezone aware so I can convert it properly? In this case I know the timezone, but what if I did not know?
The error is because you're not including the time part in the format string. Do that:
datetime.strptime('2019-10-07T06:09:28.984Z', '%Y-%m-%dT%H:%M:%S.%f%z')
This results in:
datetime.datetime(2019, 10, 7, 6, 9, 28, 984000, tzinfo=datetime.timezone.utc)
If you want to convert this to a local timezone, do that:
from pytz import timezone
dt = datetime.strptime(strdate, '%Y-%m-%dT%H:%M:%S.%f%z')
local_dt = dt.astimezone(timezone('Asia/Tokyo'))
Demo: https://repl.it/repls/RotatingSqueakyCertifications
import datetime
strdate = '2019-10-07T06:09:28.984Z'
dt=datetime.datetime.strptime(strdate, "%Y-%m-%dT%H:%M:%S.%fZ")
print(dt)
# output - 2019-10-07 06:09:28.984000
I am very new to python (2.6.7)
How would I be able convert the output of:
todays_date = datetime.datetime.today()
print todays_date
yesterday = todays_date - timedelta(days=1)
print yesterday
Which yields to:
2015-11-22 23:44:45.166081
2015-11-21 23:44:45.166081
From this, I want to change the format of yesterday to just 21/11/2015
How could this be done?
If you have a datetime object then to get a date part, call its .date() method:
just_date = some_datetime.date()
Note: don't confuse a date object such date.today() and its text representation such as str(date.today()):
>>> from datetime import date, timedelta
>>> yesterday = date.today() - timedelta(1)
>>> yesterday
datetime.date(2015, 11, 22)
>>> str(yesterday)
'2015-11-22'
>>> print yesterday
2015-11-22
>>> yesterday.strftime('%d/%m/%Y')
'22/11/2015'
As an input to an API request I need to get yesterday's date as a string in the format YYYY-MM-DD. I have a working version which is:
yesterday = datetime.date.fromordinal(datetime.date.today().toordinal()-1)
report_date = str(yesterday.year) + \
('-' if len(str(yesterday.month)) == 2 else '-0') + str(yesterday.month) + \
('-' if len(str(yesterday.day)) == 2 else '-0') + str(yesterday.day)
There must be a more elegant way to do this, interested for educational purposes as much as anything else!
You Just need to subtract one day from today's date. In Python datetime.timedelta object lets you create specific spans of time as a timedelta object.
datetime.timedelta(1) gives you the duration of "one day" and is subtractable from a datetime object. After you subtracted the objects you can use datetime.strftime in order to convert the result --which is a date object-- to string format based on your format of choice:
>>> from datetime import datetime, timedelta
>>> yesterday = datetime.now() - timedelta(1)
>>> type(yesterday)
>>> datetime.datetime
>>> datetime.strftime(yesterday, '%Y-%m-%d')
'2015-05-26'
Note that instead of calling the datetime.strftime function, you can also directly use strftime method of datetime objects:
>>> (datetime.now() - timedelta(1)).strftime('%Y-%m-%d')
'2015-05-26'
As a function:
from datetime import datetime, timedelta
def yesterday(frmt='%Y-%m-%d', string=True):
yesterday = datetime.now() - timedelta(1)
if string:
return yesterday.strftime(frmt)
return yesterday
example:
In [10]: yesterday()
Out[10]: '2022-05-13'
In [11]: yesterday(string=False)
Out[11]: datetime.datetime(2022, 5, 13, 12, 34, 31, 701270)
An alternative answer that uses today() method to calculate current date and then subtracts one using timedelta(). Rest of the steps remain the same.
https://docs.python.org/3.7/library/datetime.html#timedelta-objects
from datetime import date, timedelta
today = date.today()
yesterday = today - timedelta(days = 1)
print(today)
print(yesterday)
Output:
2019-06-14
2019-06-13
>>> import datetime
>>> datetime.date.fromordinal(datetime.date.today().toordinal()-1).strftime("%F")
'2015-05-26'
Calling .isoformat() on a date object will give you YYYY-MM-DD
from datetime import date, timedelta
(date.today() - timedelta(1)).isoformat()
I'm trying to use only import datetime based on this answer.
import datetime
oneday = datetime.timedelta(days=1)
yesterday = datetime.date.today() - oneday