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python json date object to python datetime

I have a JSON object with a date that returns
print row['ApplicationReceivedDateTime']
/Date(1454475600000)/
how do I process this using the pythons datetime module?
print type(row['ApplicationReceivedDateTime'])
returns <type 'unicode'>
print repr(row['ApplicationReceivedDateTime'])
returns u'/Date(1454475600000)/'

That looks like milliseconds. Try dividing by 1000.
import datetime as dt
>>> dt.datetime.fromtimestamp(1454475600000 / 1000)
datetime.datetime(2016, 2, 2, 21, 0)
If the date is in the string format per your question, extract the numeric portion using re.
date = '/Date(1454475600000)/'
>>> dt.datetime.fromtimestamp(int(re.findall(r"\d+", date)[0]) / 1000)
datetime.datetime(2016, 2, 2, 21, 0)

You probably want
datetime.datetime.strptime(string_date, "%Y-%m-%d %H:%M:%S.%f")
And the values of Year, Month, Day, Hour, Minute, Second and F, for that you can write a manual function for that like this
def generate_date_time_str(date_str):
"""Login to parse the date str"""
return date_str
the date_str will look link this
"%Y-%m-%d %H:%M:%S.%f"
There is no python module directly convert any random date str to DateTime object

You can use re to get the integer value and then use datetime.datetime.fromtimestamp to get the date value:
from datetime import datetime
import re
string_time = row['ApplicationReceivedDateTime']
parsed_time = int(re.search('\((\d+)\)', string_time)[1]) / 1e3 #1e3 == 1000
rcvd_date = datetime.fromtimestamp(parsed_time)
print(rcvd_date.strftime('%Y-%m-%d %H:%M:%S'))
Prints:
'2016-02-03 05:00:00'

Get file modification date in MM-DD-YYYY format without time [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]

Convert a datetame object to the correct date (MM-DD-YYYY to DD-MM-YYYY)

I have parsed a date and stored it as a datetime object. The date was written in the format MM-DD-YYYY instead of DD-MM-YYYY when it was parsed. What would be the easiest way to convert the object to the correct date?

You can swap out values with the datetime.datetime.replace() method, provided the day value is within the range 1-12, of course:
dt = dt.replace(month=dt.day, day=dt.month)
The method returns a new datetime instance.
Demo:
>>> from datetime import datetime
>>> dt = datetime(2015, 2, 11)
>>> dt
datetime.datetime(2015, 2, 11, 0, 0)
>>> dt.replace(month=dt.day, day=dt.month)
datetime.datetime(2015, 11, 2, 0, 0)

Try this out
>>> import datetime
>>> d = datetime.datetime.strptime('2011-06-09', '%Y-%m-%d')
>>> d.strftime('%d-%m-%Y')
'09-06-2011'
Not working? Let me know :)

Cannot use datetime.datetime.strptime, got an AttributeError instead

I have two strings in python.
time_array1='09-JAN-2014 01:19'
time_array2='09-JAN-2014 01:01'
I need to find the time difference and I am doing:
print time_array1
print time_array2
FMT = '%d-%m-%Y %H:%M'
datetime_object1= datetime.datetime.strptime(time_array1, FMT)
print datetime_object1
datetime_object2= datetime.datetime.strptime(time_array2, FMT)
print datetime_object2
diff=datetime_object1 - datetime_object2
print diff
but I am getting the following error:
datetime_object1= datetime.datetime.strptime(time_array1, FMT)
AttributeError: type object 'datetime.datetime' has no attribute 'strptime'
Is there any alternative way through which I can do it. It seems the python library doesn't have strptime attribute.

The strptime method was added in python 2.5; if you are using an older version use the following code instead:
import datetime, time
datetime_object1 = datetime.datetime(*time.strptime(time_array1, FMT)[:6])
Your months are abbreviations, use the %b to parse that instead of %m.
Demo:
>>> import datetime, time
>>> time_array1='09-JAN-2014 01:19'
>>> time_array2='09-JAN-2014 01:01'
>>> FMT = '%d-%b-%Y %H:%M'
>>> datetime.datetime(*time.strptime(time_array1, FMT)[:6])
datetime.datetime(2014, 1, 9, 1, 19)
>>> datetime.datetime(*time.strptime(time_array2, FMT)[:6])
datetime.datetime(2014, 1, 9, 1, 1)

Date Time Formats in Python

What are these date-time formats? I need to convert them to the same format, to check if they are the same. These are just two coming from a separate data source, so I need to find a way to make them the same format. Any ideas?
2013-07-12T07:00:00Z
2013-07-10T11:00:00.000Z
Thanks in advance

That extra .000 is micro seconds.
This will convert a date string of a format to datetime object.
import datetime
d1 = datetime.datetime.strptime("2013-07-12T07:00:00Z","%Y-%m-%dT%H:%M:%SZ")
d2 = datetime.datetime.strptime("2013-07-10T11:00:00.000Z","%Y-%m-%dT%H:%M:%S.%fZ")
Then convert them into any format depending on your requirement, by using:
new_format = "%Y-%m-%d"
d1.strftime(new_format)

perhaps use .isoformat()
string in ISO 8601 format, YYYY-MM-DDTHH:MM:SS[.mmmmmm][+HH:MM]
>>> import datetime
>>> datetime.datetime.utcnow().isoformat() + "Z"
'2013-07-11T22:26:51.564000Z'
>>>
Z specifies "zulu" time or UTC.
You can also add the timezone component by making your datetime object timezone aware by applying the appropriate tzinfo object. With the tzinfo applied the .isoformat() method will include the appropriate utc offset in the output:
>>> d = datetime.datetime.utcnow().replace(tzinfo=datetime.timezone.utc)
>>> d.isoformat()
'2019-11-11T00:52:43.349356+00:00'
You can remove the microseconds by change the microseconds value to 0:
>>> no_ms = d.replace(microsecond=0)
>>> no_ms.isoformat()
'2019-11-11T00:52:43+00:00'
Also, as of python 3.7 the .fromisoformat() method is available to load an iso formatted datetime string into a python datetime object:
>>> datetime.datetime.fromisoformat('2019-11-11T00:52:43+00:00')
datetime.datetime(2019, 11, 11, 0, 52, 43, tzinfo=datetime.timezone.utc)
http://www.ietf.org/rfc/rfc3339.txt

you can try to trim the string
data = "2019-10-22T00:00:00.000-05:00"
result1 = datetime.datetime.strptime(data[0:19],"%Y-%m-%dT%H:%M:%S")
result2 = datetime.datetime.strptime(data[0:23],"%Y-%m-%dT%H:%M:%S.%f")
result3 = datetime.datetime.strptime(data[0:9], "%Y-%m-%d")

use datetime module.
For a variable
import datetime
def convertDate(d):
new_date = datetime.datetime.strptime(d,"%Y-%m-%dT%H:%M:%S.%fZ")
return new_date.date()
convertDate("2019-12-23T00:00:00.000Z")
you can change the ".date()" to ".year", ".month", ".day" etc...
Output: # is now a datetime object
datetime.date(2019, 12, 23)
For a DataFrame column, use apply()
df['new_column'] = df['date_column'].apply(convertDate)

* Short and best way:
str(datetime.datetime.now()).replace(' ','T')
or
str(datetime.datetime.now()).replace(' ','T') + "Z"