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convert 2021-01-18T11:18:10.833876+00:00 to datetime python

I am trying to convert the following string to datetime in python. After referring to datetime.strptime(‘2017-01-12T14:12:06.000-0500’,'%Y-%m-%dT%H:%M:%S.%f%Z') I am trying the below mentioned format.
config_current_ts = datetime.strptime(internal_config["timestamp_str"], "%Y:%m:%dT%H:%M:%S.%f%z")
Yet I get an error stating :
ValueError: time data '2021-01-18T11:18:10.833876+00:00' does not
match format '%Y:%m:%dT%H:%M:%S.%f%z'
I am using python3.7.4. Can someone tell me how to convert this to datetime? I want to basically compare the string and current time to see which is ahead.

With Python 3.7+, use fromisoformat - since you have ISO 8601 format, it is appropriate and as a benefit also more efficient. Ex:
from datetime import datetime
s = '2021-01-18T11:18:10.833876+00:00'
dt = datetime.fromisoformat(s)
print(dt)
# 2021-01-18 11:18:10.833876+00:00
print(repr(dt))
# datetime.datetime(2021, 1, 18, 11, 18, 10, 833876, tzinfo=datetime.timezone.utc)

the format string should be '%Y-%m-%dT%H:%M:%S.%f%z'
>>> from datetime import datetime
>>> d = '2021-01-18T11:18:10.833876+00:00'
>>> datetime.strptime(d, '%Y-%m-%dT%H:%M:%S.%f%z')
datetime.datetime(2021, 1, 18, 11, 18, 10, 833876, tzinfo=datetime.timezone.utc)

Python: How to Drop Hours from a Datetime? [duplicate]

I have a date string and want to convert it to the date type:
I have tried to use datetime.datetime.strptime with the format that I want but it is returning the time with the conversion.
when = alldates[int(daypos[0])]
print when, type(when)
then = datetime.datetime.strptime(when, '%Y-%m-%d')
print then, type(then)
This is what the output returns:
2013-05-07 <type 'str'>
2013-05-07 00:00:00 <type 'datetime.datetime'>
I need to remove the time: 00:00:00.

print then.date()
What you want is a datetime.date object. What you have is a datetime.datetime object. You can either change the object when you print as per above, or do the following when creating the object:
then = datetime.datetime.strptime(when, '%Y-%m-%d').date()

If you need the result to be timezone-aware, you can use the replace() method of datetime objects. This preserves timezone, so you can do
>>> from django.utils import timezone
>>> now = timezone.now()
>>> now
datetime.datetime(2018, 8, 30, 14, 15, 43, 726252, tzinfo=<UTC>)
>>> now.replace(hour=0, minute=0, second=0, microsecond=0)
datetime.datetime(2018, 8, 30, 0, 0, tzinfo=<UTC>)
Note that this returns a new datetime object -- now remains unchanged.

>>> print then.date(), type(then.date())
2013-05-07 <type 'datetime.date'>

To convert a string into a date, the easiest way AFAIK is the dateutil module:
import dateutil.parser
datetime_object = dateutil.parser.parse("2013-05-07")
It can also handle time zones:
print(dateutil.parser.parse("2013-05-07"))
>>> datetime.datetime(2013, 5, 7, 1, 12, 12, tzinfo=tzutc())
If you have a datetime object, say:
import pytz
import datetime
now = datetime.datetime.now(pytz.UTC)
and you want chop off the time part, then I think it is easier to construct a new object instead of "substracting the time part". It is shorter and more bullet proof:
date_part datetime.datetime(now.year, now.month, now.day, tzinfo=now.tzinfo)
It also keeps the time zone information, it is easier to read and understand than a timedelta substraction, and you also have the option to give a different time zone in the same step (which makes sense, since you will have zero time part anyway).

For me, I needed to KEEP a timetime object because I was using UTC and it's a bit of a pain. So, this is what I ended up doing:
date = datetime.datetime.utcnow()
start_of_day = date - datetime.timedelta(
hours=date.hour,
minutes=date.minute,
seconds=date.second,
microseconds=date.microsecond
)
end_of_day = start_of_day + datetime.timedelta(
hours=23,
minutes=59,
seconds=59
)
Example output:
>>> date
datetime.datetime(2016, 10, 14, 17, 21, 5, 511600)
>>> start_of_day
datetime.datetime(2016, 10, 14, 0, 0)
>>> end_of_day
datetime.datetime(2016, 10, 14, 23, 59, 59)

If you specifically want a datetime and not a date but want the time zero'd out you could combine date with datetime.min.time()
Example:
datetime.datetime.combine(datetime.datetime.today().date(),
datetime.datetime.min.time())

You can use simply pd.to_datetime(then) and pandas will convert the date elements into ISO date format- [YYYY-MM-DD].
You can pass this as map/apply to use it in a dataframe/series too.

You can usee the following code:
week_start = str(datetime.today() - timedelta(days=datetime.today().weekday() % 7)).split(' ')[0]

normalizing JSON datestrings to UTC python

I have an important test that says "Calculate users that logged in during the month of April normalized to the UTC timezone."
Items look as such:
[ {u'email': u' ybartoletti#littel.biz',
u'login_date': u'2014-05-08T22:30:57-04:00'},
{u'email': u'woodie.crooks#kozey.com',
u'login_date': u'2014-04-25T13:27:48-08:00'},
]
It seems to me that an item like 2014-04-13T17:12:20-04:00 means "April 13th, 2014, at 5:12:20 pm, 4 hours behind UTC". Then I just use strptime to convert to datetime (Converting JSON date string to python datetime), and subtract a timedelta of however many hours I get from a regex that grabs the end of string? I feel this way because some have a + at the end instead of -, like 2014-05-07T00:30:06+07:00
Thank you

It is probably best to use the dateutil.parser.parse and pytz packages for this purpose. This will allow you to parse a string and convert it to a datetime object with UTC timezone:
>>> s = '2014-05-08T22:30:57-04:00'
>>> import dateutil.parser
>>> import pytz
>>> pytz.UTC.normalize(dateutil.parser.parse(s))
datetime.datetime(2014, 5, 9, 2, 30, 57, tzinfo=<UTC>)

You can use arrow to easily parse date with time zone.
>>>import arrow
>>> a = arrow.get('2014-05-08T22:30:57-04:00').to('utc')
>>> a
<Arrow [2014-05-09T02:30:57+00:00]>
Get a datetime object or timestamp:
>>> a.datetime
datetime.datetime(2014, 5, 9, 2, 30, 57, tzinfo=tzutc())
>>> a.naive
datetime.datetime(2014, 5, 9, 2, 30, 57)
>>> a.timestamp
1399602657

The following solution should be faster and avoids importing external libraries. The downside is that it will only work if the date strings are all guaranteed to have the specified format. If that's not the case, then I would prefer Simeon's solution, which lets dateutil.parser.parse() take care of any inconsistencies.
import datetime as dt
def parse_date(datestr):
diff = dt.timedelta(hours=int(datestr[20:22]), minutes=int(datestr[23:]))
if datestr[19] == '-':
return dt.datetime.strptime(datestr[:19], '%Y-%m-%dT%H:%M:%S') - diff
return dt.datetime.strptime(datestr[:19], '%Y-%m-%dT%H:%M:%S') + diff

Python: convert complicated date and time string to timestamp

I want to know how to convert this date format
"Thu 21st Aug '14, 4:58am"
to a timestamp with Python?
Another format that I need to convert:
"Yesterday, 7:22am"
I tried parse util without success...

If you haven't done so already, have a look at the parse function in dateutils.parser for parsing strings representing dates...
>>> from dateutil.parser import parse
>>> dt = parse("Thu 21st Aug '14, 4:58am")
>>> dt
datetime.datetime(2014, 8, 21, 4, 58)
...and then to convert a datetime object to a timestamp, you can do the following:
>>> import time
>>> import datetime
>>> time.mktime(dt.timetuple())
1408593480.0
As side remark, parse is a useful function which can recognise a huge range of different date formats. However it's sometimes too helpful and sees dates where perhaps a date is not intended:
>>> parse("14, m 23")
datetime.datetime(2014, 8, 23, 0, 14)
If you also want to parse expressions such as "Yesterday, 7:22am", you could do one of two things:
Replace "yesterday", "yester-day", "yday" and other variations with "25/08/2014" (or another appropriate date) and then use parse on the new string.
Use another library to parse the string. parsedatetime is one option...
Here's parsedatetime in action on your example:
>>> import parsedatetime.parsedatetime as pdt
>>> p = pdt.Calendar()
>>> d = p.parse("Yesterday, 7:22am")
>>> d
((2014, 8, 25, 7, 22, 0, 0, 237, -1), 3)
To turn this date representation d into a datetime object, you can unpack the tuple like so:
>>> dt = datetime.datetime(*d[0][:7])
>>> dt
datetime.datetime(2014, 8, 25, 7, 22)
Now dt can be easily converted to a timestamp in the way described above.

You can use this:
a = "Thu 21st Aug '14, 4:58am"
datetime.datetime.strptime(a, '%a %dst %b \'%y, %H:%M%p')

Convert UTC time to python datetime

I have numerous UTC time stamps in the following format:
2012-04-30T23:08:56+00:00
I want to convert them to python datetime objects but am having trouble.
My code:
for time in data:
pythondata[i]=datetime.strptime(time,"%y-%m-%dT%H:%M:%S+00:00")
I get the following error:
ValueError: time data '2012-03-01T00:05:55+00:00' does not match format '%y-%m-%dT%H:%M:%S+00:00'
It looks like I have the proper format, so why doesn't this work?

Change the year marker in your time format string to %Y:
time = '2012-03-01T00:05:55+00:00'
datetime.strptime(time, "%Y-%m-%dT%H:%M:%S+00:00")
# => datetime.datetime(2012, 3, 1, 0, 5, 55)
See strftime() and strptime() behavior.

I highly recommend python-dateutil library, it allows conversion of multiple datetime formats from raw strings into datetime objects with/without timezone set
>>> from dateutil.parser import parse
>>> parse('2012-04-30T23:08:56+00:00')
datetime.datetime(2012, 4, 30, 23, 8, 56, tzinfo=tzutc())