I use connect the sqlite3 of chrome 'History' file for analysis. But there are data for last_visit_time, which is timestamp, but I don't know how to change it to real time.
I test like this:
2012-04-18 23:22:11.084300 (utctime)
2012-04-19 01:22:11.084300 (myPC time)
12,979,264,931,952,304 (the time stamp in the data)
I printed the utctime and clicked a website, almost at the same time. So I get the statistic like above. 12,979,264,931,952,304 is the long int, so normal way to convert is not possible.
How can I convert the timestamp to a date?
The timestamp it stores is the number of microseconds since midnight UTC on January 1st, 1601. To convert this to the current time, you can add the number of microseconds you have to the epoch date, with help from this answer, like this:
>>> import datetime
>>> epoch_start = datetime.datetime(1601, 1, 1)
>>> delta = datetime.timedelta(microseconds=12979264931952304)
>>> epoch_start + delta
datetime.datetime(2012, 4, 18, 23, 22, 11, 952304)
To convert to your local timezone, you can use this method (note that I am currently UTC - 4, while it looks like you are UTC + 2):
>>> from dateutil import tz
>>> from_zone = tz.tzutc()
>>> to_zone = tz.tzlocal()
>>> utc_time = (epoch_start + delta).replace(tzinfo=from_zone)
>>> utc_time.astimezone(to_zone)
datetime.datetime(2012, 4, 18, 19, 22, 11, 952304, tzinfo=tzlocal())
Related
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]
I have time in J2000 format, ie. seconds after noon on 1-1-2000, which I want to convert to UTC time in an array with the format [year-month-day hour:min:sec.millisec]. Is there some function in AstroPy or something similar to do the conversion?
Input:
Time in J2000: 559889620.293 seconds
Desired Output:
Time in UTC: 2017-09-28 16:53:40.293
If you want to use astropy you could use Time and TimeDelta from astropy.time:
>>> from astropy.time import TimeDelta, Time
>>> from astropy import units as u
>>> (Time(2000, format='jyear') + TimeDelta(559889620.293*u.s)).iso
'2017-09-28 16:53:35.293'
The Time(2000, format='jyear') is a good alternative if you don't want to remember what the baseline for julian dates (noon on 1.1.2000) is.
You can simply add a timedelta to a base datetime.
import datetime
datetime.datetime(2000, 1, 1, 12, 0) + datetime.timedelta(seconds=559889620.293)
returns:
datetime.datetime(2017, 9, 28, 16, 53, 40, 293000)
The offset is constant, so you can just add it:
>>> OFFSET = datetime(2000,1,1,12) - datetime(1970,1,1)
>>> datetime.utcfromtimestamp(559889620.293) + OFFSET
datetime.datetime(2017, 9, 28, 16, 53, 40, 293000)
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
What is the difference between the 2 statements:
import datetime
print datetime.datetime.now()
datetime.datetime(2015, 1, 28, 12, 32, 9, 762118)
from datetime import *
>> datetime.time(datetime.now())
datetime.time(12, 33, 3, 693084)
Actually I want to compare TimeField of a django model with the current day's 1 hour less time. My code snippet for the same:
Mymodel.objects.filter(
follow_up_date=datetime.datetime.now().date,
# commented now
# preferred_time__lt=(
# datetime.datetime.now() - datetime.timedelta(hours=1)),
preferred_time__lt=(datetime.time(datetime.now()) - datetime.timedelta(hours=1)),
)
Mymodel:
class Mymodel(models.Model):
follow_up_date = models.DateField("Follow up Date",null=True,blank=True)
preferred_time = models.TimeField(_("Preferred time"),default=now,
null=True,blank=True)
I am trying to extract all the instances which are scheduled for the day, whose preferred time has elapsed just 1 hour back. Which should be the correct filter for the 'preferred_time'? I got wrong results for the commented code. I am not clear.
This is a cron job i.e management command to be run every 1 hour in django
In the first instance:
datetime.datetime(2015, 1, 28, 12, 32, 9, 762118)
You have a datetime object. It has both the date (first three numbers) and the time (last four numbers).
The second object you mention:
datetime.time(12, 33, 3, 693084)
This is just the time component.
To compare a TimeField, you need just the time component; for a DateField, just the date component.
In your code, you have the following datetime.datetime.now().date this is just the name of the built-in function date. You need to call it:
>>> datetime.datetime.now().date
<built-in method date of datetime.datetime object at 0xb74ac9b0>
>>> datetime.datetime.now().date()
datetime.date(2015, 1, 28)
You also cannot do datetime.time(datetime.datetime.now()), datetime.time() is a constructor method, it is not a way to covert other objects.
You also cannot subtract timedelta from a time object:
To get the correct result, you need to subtract one hour from the datetime object, then convert it to time:
>>> (datetime.datetime.now() - datetime.timedelta(hours=1)).time()
datetime.time(9, 27, 16, 93746)
In the end, your filter would look like:
filter_date = datetime.datetime.now().date()
filter_time = (datetime.datetime.now() - datetime.timedelta(hours=1)).time()
Mymodel.objects.filter(follow_up_date=filter_date,
preferred_time__lt=filter_time)
datetime.now() given date and time information.
datetime.time() give only time information.
e.g
>>> from datetime import *
>>> datetime.now()
datetime.datetime(2015, 1, 28, 12, 52, 35, 164500)
>>> datetime.time(datetime.now())
datetime.time(12, 52, 41, 97521)
>>>
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]