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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]

Substract current time to -GMT in python

I have a time which is 13:11:06 and i want to -GMT (i.e -0530). I can minus it by simply doing -5 by splitting the string taking the first digit (convert to int) and then minus it and then re-join. But then i get it in a format which is 8:11:06 which is not right as it should be 08:11:06, secondly its a lengthy process. Is there a easy way to get my time in -GMT format (08:11:06)
This is what i did to get -GMT time after getting the datetime
timesplithour = int(timesplit[1]) + -5
timesplitminute = timesplit[2]
timesplitseconds = timesplit[3]
print timesplithour
print timesplitminute
print timesplitseconds
print timesplithour + ":" + timesplitminute + ":" + timesplitseconds

You could use Python's datatime library to help you as follows:
import datetime
my_time = "13:11:06"
new_time = datetime.datetime.strptime("2016 " + my_time, "%Y %H:%M:%S") - datetime.timedelta(hours=5, minutes=30)
print new_time.strftime("%H:%M:%S")
This would print:
07:41:06
First it converts your string into a datetime object. It then creates a timedelta object allowing you to subtract 5 hours 30 minutes from the datetime object. Finally it uses strftime to format the resulting datetime into a string in the same format.

Use the datetime module:
from datetime import datetime, timedelta
dt = datetime.strptime('13:11:06', '%H:%M:%S')
time_gmt = (dt - timedelta(hours=5, minutes=30)).time()
print(time_gmt.hour)
print(time_gmt.minute)
print(time_gmt.second)
s = time_gmt.strftime('%H:%M:%S')
print(s)
Output
7
41
6
07:41:06
Note that this subtracts 5 hours and 30 minutes as initially mentioned in the question. If you really only want to subtract 5 hours, use timedelta(hours=5).

You can use datetimes timedelta.
print datetime.datetime.today()
>>> datetime.datetime(2016, 3, 3, 10, 45, 6, 270711)
print datetime.datetime.today() - datetime.timedelta(days=3)
>>> datetime.datetime(2016, 2, 29, 10, 45, 8, 559073)
This way you can subtract easily

Assuming the time is a datetime instance
import datetime as dt
t = datetime(2015,12,31,13,11,06)
#t.time() # gives time object. ie no date information
offset = dt.timedelta(hours=5,minutes=30) # or hours=5.5
t -= offset
t.strftime(("%H:%M:%S") # output as your desired string
#'18:41:06'

If the object is datetime and you don't care about DST, the simplest thing you can do is,
In [1]: from datetime import datetime
In [2]: curr = datetime.now()
In [3]: curr
Out[3]: datetime.datetime(2016, 3, 3, 9, 57, 31, 302231)
In [4]: curr.utcnow()
Out[4]: datetime.datetime(2016, 3, 3, 8, 57, 57, 286956)

Parsing non-zero padded timestamps in Python

I want to get datetimes from timestamps like the following :3/1/2014 9:55 with datetime.strptime, or something equivalent.
The month, day of month, and hour is not zero padded, but there doesn't seem to be a formatting directive listed here that is able to parse this automatically.
What's the best approach to do so? Thanks!

strptime is able to parse non-padded values. The fact that they are noted as being padded in the formatting codes table applies to strftime's output. So you can just use
datetime.strptime(datestr, "%m/%d/%Y %H:%M")

strptime isdo not require 0-padded values. See example below
datetime.strptime("3/1/2014 9:55", "%m/%d/%Y %H:%M")
output: datetime.datetime(2014, 3, 1, 9, 55)

The non-pattern way is use dateutil.parse module, it lets to parse the common date formats, even if you don't know what it is using currently
Ex:
>>> import dateutil.parser
>>>
>>> utc_time = '2014-08-13T00:00:00'
>>> verbose_time = '13-Aug-2014'
>>> some_locale = '3/1/2014 9:55'
>>> dateutil.parser.parse(utc_time)
datetime.datetime(2014, 8, 13, 0, 0)
>>> dateutil.parser.parse(verbose_time)
datetime.datetime(2014, 8, 13, 0, 0)
>>> dateutil.parser.parse(some_locale)
datetime.datetime(2014, 3, 1, 9, 55)

Just in case this answer helps someone else -- I came here thinking I had a problem with zero padding, but it was actually to do with 12:00 vs 00:00 and the %I formatter.
The %I formatter is meant to match 12-hour-clock hours, optionally zero-padded. But depending on your data source, you might get data that says that midnight or midday is actually zero, eg:
>>> datetime.strptime('2015/01/01 0:12am', "%Y/%m/%d %I:%M%p")
ValueError: time data '2015/01/01 0:12am' does not match format '%Y/%m/%d %I:%M'
What strptime actually wanted was a 12, not a zero:
>>> datetime.strptime('2015/01/01 12:12am', "%Y/%m/%d %I:%M%p")
datetime.datetime(2015, 1, 1, 0, 12)
But we don't always control our data sources! My solution for this edge case was to catch the exception, try parsing it with a %H, with a quick check that we are in the edge case we think we are in.
def get_datetime(string):
try:
timestamp = datetime.strptime(string, "%m/%d/%Y %I:%M%p")
except ValueError:
# someone used zero for midnight?
timestamp = datetime.strptime(string, "%m/%d/%Y %H:%M%p")
assert string.lower().endswith('am')
assert timestamp.hour == 0
return timestamp

You can see the strftime document here，but in fact they aren't all working well in all platforms，for instance，%-d，%-m don't work on win7 by python 2.7，so you can accomplish like this
>>> date_str = '{d.year}-{d.month}-{d.day}'.format(d=datetime.datetime.now())
>>> print(date_str)
2016-5-23

Get date object for the first/last day of the current year

I need to get date objects for the first and last day in the current year.
Currently I'm using this code which works fine, but I'm curious if there's a nicer way to do it; e.g. without having to specify the month/day manually.
from datetime import date
a = date(date.today().year, 1, 1)
b = date(date.today().year, 12, 31)

The only real improvement that comes to mind is to give your variables more descriptive names than a and b.

from datetime import datetime
starting_day_of_current_year = datetime.now().date().replace(month=1, day=1)
ending_day_of_current_year = datetime.now().date().replace(month=12, day=31)

There is nothing in the python library but there are external libraries that wrap this functionality up. For example, pandas has a timeseries library, with which you can do:
from datetime import date
from pandas.tseries import offsets
a = date.today() - offsets.YearBegin()
b = date.today() + offsets.YearEnd()
Whilst pandas is overkill if all you want is year begin and year end functionality, it also has support for a lot of other high level concepts such as business days, holiday calendars, month/quarter/year offsets: http://pandas.pydata.org/pandas-docs/stable/timeseries.html#dateoffset-objects

You'll have some funky stuff going on if you happen to be running this late on New Year's Eve and the two calls to today() cross the year boundary. It's safer to do this:
from datetime import date
epoch_year = date.today().year
year_start = date(epoch_year, 1, 1)
year_end = date(epoch_year, 12, 31)

import datetime
year = 2016
first_day_of_year = datetime.date.min.replace(year = year)
last_day_of_year = datetime.date.max.replace(year = year)
print(first_day_of_year, last_day_of_year)
duh.
What if we want to get the exact time the year begins or ends? Same thing. Replace datetime.date with datetime.datetime, and there you go, you got the first last day of year in datetime.datetime format.
To make this even fancier, wrap the whole thing in a function:
import datetime
def year_range(year, datetime_o = datetime.date):
return (
datetime_o.min.replace(year = year),
datetime_o.max.replace(year = year)
)
print(year_range(2016, datetime.date))
print(year_range(2016, datetime.datetime))
Output:
(datetime.date(2016, 1, 1), datetime.date(2016, 12, 31))
(datetime.datetime(2016, 1, 1, 0, 0), datetime.datetime(2016, 12, 31, 23, 59, 59, 999999))

use relative timedelta and substract from date object
from dateutil.relativedelta import relativedelta
import datetime
date = datetime.date.today()
fistOfYear = date - relativedelta(years=0, month=1, day=1)

one genius way to find out first and last day of year is code below
this code works well even for leap years
first_day=datetime.date(year=i,month=1, day=1)
first_day_of_next_year=first_day.replace(year=first_day.year+1,month=1, day=1)
last_day=first_day_of_next_year-jdatetime.timedelta(days=1)