What is the object used in Python to specify date (and time) in Python?
For instance, to create an object that holds a given date and time, (let's say '05/10/09 18:00').
As per S.Lott's request, so far I have:
class Some:
date =
I stop there. After the "=" sign for, I realize I didn't knew what the right object was ;)
Simple example:
>>> import datetime
# 05/10/09 18:00
>>> d = datetime.datetime(2009, 10, 5, 18, 00)
>>> print d.year, d.month, d.day, d.hour, d.second
2009 10 5 18 0
>>> print d.isoformat(' ')
2009-10-05 18:00:00
>>>
Nick D has the official way of handling your problem. If you want to pass in a string like you did in your question, the dateutil module (http://labix.org/python-dateutil) has excellent support for that kind of thing.
For examples, I'm going to copy and paste from another answer I gave a while back now:
Simple example:
>>> parse("Thu Sep 25 2003")
datetime.datetime(2003, 9, 25, 0, 0)
>>> parse("Sep 25 2003")
datetime.datetime(2003, 9, 25, 0, 0)
>>> parse("Sep 2003", default=DEFAULT)
datetime.datetime(2003, 9, 25, 0, 0)
>>> parse("Sep", default=DEFAULT)
datetime.datetime(2003, 9, 25, 0, 0)
>>> parse("2003", default=DEFAULT)
datetime.datetime(2003, 9, 25, 0, 0)
To ambigous:
>>> parse("10-09-2003")
datetime.datetime(2003, 10, 9, 0, 0)
>>> parse("10-09-2003", dayfirst=True)
datetime.datetime(2003, 9, 10, 0, 0)
>>> parse("10-09-03")
datetime.datetime(2003, 10, 9, 0, 0)
>>> parse("10-09-03", yearfirst=True)
datetime.datetime(2010, 9, 3, 0, 0)
To all over the board:
>>> parse("Wed, July 10, '96")
datetime.datetime(1996, 7, 10, 0, 0)
>>> parse("1996.07.10 AD at 15:08:56 PDT", ignoretz=True)
datetime.datetime(1996, 7, 10, 15, 8, 56)
>>> parse("Tuesday, April 12, 1952 AD 3:30:42pm PST", ignoretz=True)
datetime.datetime(1952, 4, 12, 15, 30, 42)
>>> parse("November 5, 1994, 8:15:30 am EST", ignoretz=True)
datetime.datetime(1994, 11, 5, 8, 15, 30)
>>> parse("3rd of May 2001")
datetime.datetime(2001, 5, 3, 0, 0)
>>> parse("5:50 A.M. on June 13, 1990")
datetime.datetime(1990, 6, 13, 5, 50)
Take a look at the documentation for it here:
http://labix.org/python-dateutil#head-c0e81a473b647dfa787dc11e8c69557ec2c3ecd2
Look at the datetime module; there are datetime, date and timedelta class definitions.
>>> import datetime
>>> datetime.datetime.strptime('05/10/09 18:00', '%d/%m/%y %H:%M')
datetime.datetime(2009, 10, 5, 18, 0)
>>> datetime.datetime.today()
datetime.datetime(2009, 10, 5, 21, 3, 55, 827787)
So, you can either use format string to convert to datetime.datetime object or if you're particularly looking at today's date could use today() function.
Related
I need to set the time value of detatime objects to 00:00, so I can easily compare two 'dates'
Now I do:
extracted_start_date = datetime.strptime(extracted_start_date.strftime('%Y-%m-%d'), '%Y-%m-%d')
so I have my datetime object
$ datetime.datetime(2021, 3, 18, 11, 13, 53, 782088),
I extract the date string strftime('%Y-%m-%d')
$ '2021-03-18'
and put this back into a datetime object now wch has time 00:00
$ datetime.datetime(2021, 3, 18, 0, 0)
This seems quite elaborate, is there a more efficient way, or is it OK like this?
datetime.datetime objects have method date which return datetime.date instance and these might be used for day-based comparison, consider following example:
import datetime
d1 = datetime.datetime(2021, 3, 17, 9, 0, 0) # yesterday 9:00
d2 = datetime.datetime(2021, 3, 18, 9, 0, 0) # today 9:00
d3 = datetime.datetime(2021, 3, 18, 12, 0, 0) # today 12:00
print(d1.date() == d2.date())
print(d2.date() == d3.date())
output:
False
True
You could use the replace method instead.
from datetime import datetime
d = datetime(2021, 3, 18, 11, 13, 53, 782088)
d.replace(hour=0, minute=0, second=0, microsecond=0)
> datetime.datetime(2021, 3, 18, 0, 0)
I want to create a mapping of the everyday of the week with its datetime object. So my dictionary should be having keys as "Monday", "Tuesday", .. (so on) so that I can get a datetime object for every day on the next(!) week.
At the moment I have a dictionary with these values:
DAYS_DATETIME_RELATIONS = {
"today": datetime.datetime.now(),
"tomorrow": datetime.datetime.now() + datetime.timedelta(days=1),
"after_tomorrow": datetime.datetime.now() + datetime.timedelta(days=2)
}
Unfortunately I cannot find any algorithmic solution for this and hope anyone of you could help me.
This can be achieved by using 2 dictionaries in the following manner:
import calendar
import datetime
days = {i: calendar.day_name[i-1] for i in range(7)}
today = datetime.datetime.now()
# using i % 7 so an arbitrary range can be used, for example
# range(7, 15) to get the week after the next week
next_week = {days[i % 7]: (today + datetime.timedelta(days=i)).date()
for i in range(7)}
print(next_week)
# {'Tuesday': datetime.date(2018, 1, 9), 'Sunday': datetime.date(2018, 1, 7),
# 'Monday': datetime.date(2018, 1, 8), 'Thursday': datetime.date(2018, 1, 11),
# 'Wednesday': datetime.date(2018, 1, 10), 'Friday': datetime.date(2018, 1, 12),
# 'Saturday': datetime.date(2018, 1, 13)}
print(next_week['Saturday'])
# 2018-01-13
Here is another way to solve your question using datetime and timedelta from datetime module:
from datetime import datetime, timedelta
def generate_dict_relation(_time, _days=0):
keys = {'Yesterday': -1, 'Today': 0, 'Tomorow': 1, 'After_tomorrow': 2}
if not _days:
return {key: _time + timedelta(days=keys.get(key, 0)) for key in keys}
else:
return {(_time + timedelta(days=_days+k)).strftime('%A'): _time + timedelta(days=_days+k)
for k in range(0, 7)}
_date_now = datetime.now()
DAYS_DATETIME_RELATIONS = {}
# get dates: yesterday, today, tomorrow and after tomorrow
DAYS_DATETIME_RELATIONS.update(generate_dict_relation(_date_now, 0))
# get dates after 7 days = 1 week
DAYS_DATETIME_RELATIONS.update(generate_dict_relation(_date_now, 7))
next_tuesday = DAYS_DATETIME_RELATIONS.get('Tuesday')
next_monday = DAYS_DATETIME_RELATIONS.get('Monday')
yesterday = DAYS_DATETIME_RELATIONS.get('Yesterday')
print('{0:%d/%m/%Y %H:%M:%S:%s} \t {1}'.format(next_tuesday, repr(next_tuesday)))
print('{0:%d/%m/%Y %H:%M:%S:%s} \t {1}'.format(next_monday, repr(next_monday)))
print('{0:%d/%m/%Y %H:%M:%S:%s} \t {1}'.format(yesterday, repr(yesterday)))
Output:
16/01/2018 10:56:26:1516096586 datetime.datetime(2018, 1, 16, 10, 56, 26, 659949)
15/01/2018 10:56:26:1516010186 datetime.datetime(2018, 1, 15, 10, 56, 26, 659949)
06/01/2018 10:56:26:1515232586 datetime.datetime(2018, 1, 6, 10, 56, 26, 659949)
One very generic way will be to create a custom iterator to return you the continuos datetime objects as:
from datetime import datetime, timedelta
class RepetetiveDate(object):
def __init__(self, day_range=7, datetime_obj=datetime.now(), jump_days=1):
self.day_range = day_range
self.day_counter = 0
self.datetime_obj = datetime_obj
self.jump_days = jump_days
self.time_deltadiff = timedelta(days=self.jump_days)
def __iter__(self):
return self
# If you are on Python 2.7
# define this function as `next(self)`
def __next__(self):
if self.day_counter >= self.day_range:
raise StopIteration
if self.day_counter != 0: # don't update for the first iteration
self.datetime_obj += self.time_deltadiff
self.day_counter += 1
return self.datetime_obj
Here, this iterator returns continuos datetime object starting from the datetime object you'll initially pass (default starts from current date).
It is using 3 optional params which you may customize as per your need:
day_range: Maximum allowed iteration for the RepetetiveDate iterator. Default value is 7.
jump_days: Integer value for jumping the number of days for the datetime object in next iteration. That means, if jump_days is equal to "2", will return datetime objects of every alternate date. To get the datetime objects of past, pass this value as negative. Default value is 1.
datetime_obj: Accepts the datetime from which date you want to start your iteration. Default value is current date.
If you are new to iterators, take a look at:
What exactly are Python's iterator, iterable, and iteration protocols?
Difference between Python's Generators and Iterators
Sample Run for upcoming dates:
>>> x = RepetetiveDate()
>>> next(x)
datetime.datetime(2018, 1, 8, 15, 55, 39, 124654)
>>> next(x)
datetime.datetime(2018, 1, 9, 15, 55, 39, 124654)
>>> next(x)
datetime.datetime(2018, 1, 10, 15, 55, 39, 124654)
Sample Run for previous dates:
>>> x = RepetetiveDate(jump_days=-1)
>>> next(x)
datetime.datetime(2018, 1, 6, 15, 55, 39, 124654)
>>> next(x)
datetime.datetime(2018, 1, 5, 15, 55, 39, 124654)
>>> next(x)
datetime.datetime(2018, 1, 4, 15, 55, 39, 124654)
How to get your desired dictionary?
Using this, you may create your dictionary using the dict comprehension as:
Dictionary of all days of week
>>> {d.strftime("%A"): d for d in RepetetiveDate(day_range=7)}
{
'Monday': datetime.datetime(2018, 1, 8, 15, 23, 16, 926364),
'Tuesday': datetime.datetime(2018, 1, 9, 15, 23, 16, 926364),
'Wednesday': datetime.datetime(2018, 1, 10, 15, 23, 16, 926364),
'Thursday': datetime.datetime(2018, 1, 11, 15, 23, 16, 926364),
'Friday': datetime.datetime(2018, 1, 12, 15, 23, 16, 926364),
'Saturday': datetime.datetime(2018, 1, 13, 15, 23, 16, 926364),
'Sunday': datetime.datetime(2018, 1, 14, 15, 23, 16, 926364)
}
Here I am using d.strftime("%A") to extract day name from the datetime object.
List of current days for next 4 weeks
>>> [d for d in RepetetiveDate(jump_days=7, day_range=4))]
[
datetime.datetime(2018, 1, 7, 16, 17, 45, 45005),
datetime.datetime(2018, 1, 14, 16, 17, 45, 45005),
datetime.datetime(2018, 1, 21, 16, 17, 45, 45005),
datetime.datetime(2018, 1, 28, 16, 17, 45, 45005)
]
One very clean way to implement this is using rrule from the dateutil library. For example:
>>> from dateutil.rrule import rrule, DAILY
>>> from datetime import datetime
>>> start_date = datetime.now()
>>> {d.strftime("%A"): d for d in rrule(freq=DAILY, count=7, dtstart=start_date)}
which will return your desired dict object:
{
'Sunday': datetime.datetime(2018, 1, 7, 17, 2, 30),
'Monday': datetime.datetime(2018, 1, 8, 17, 2, 30),
'Tuesday': datetime.datetime(2018, 1, 9, 17, 2, 30),
'Wednesday': datetime.datetime(2018, 1, 10, 17, 2, 30),
'Thursday': datetime.datetime(2018, 1, 11, 17, 2, 30),
'Friday': datetime.datetime(2018, 1, 12, 17, 2, 30),
'Saturday': datetime.datetime(2018, 1, 13, 17, 2, 30)
}
(Special thanks to Jon Clements for telling me about rrule)
I have a situation where I need to get the third latest date, i.e
INPUT :
['14-04-2001', '29-12-2061', '21-10-2019',
'07-01-1973', '19-07-2014','11-03-1992','21-10-2019']
Also , INPUT
6
14-04-2001
29-12-2061
21-10-2019
07-01-1973
19-07-2014
11-03-1992
OUTPUT : 19-07-2014
import datetime
datelist = ['14-04-2001', '29-12-2061', '21-10-2019', '07-01-1973', '19-07-2014','11-03-1992','21-10-2019' ]
for d in datelist:
x = datetime.datetime.strptime(d,'%d-%m-%Y')
print x
How can i achieve this?
You can sort the list and take the 3rd element from it.
my_list = [datetime.datetime.strptime(d,'%d-%m-%Y') for d in list]
# [datetime.datetime(2001, 4, 14, 0, 0), datetime.datetime(2061, 12, 29, 0, 0), datetime.datetime(2019, 10, 21, 0, 0), datetime.datetime(1973, 1, 7, 0, 0), datetime.datetime(2014, 7, 19, 0, 0), datetime.datetime(1992, 3, 11, 0, 0), datetime.datetime(2019, 10, 21, 0, 0)]
my_list.sort(reverse=True)
my_list[2]
# datetime.datetime(2019, 10, 21, 0, 0)
Also, as per Kerorin's suggestion, if you don't need to sort in-place and just need the 3rd element always, you can simply do
sorted(my_list, reverse=True)[2]
Update
To remove the duplicates, taking inspiration from this answer, you can do the following -
import datetime
datelist = ['14-04-2001', '29-12-2061', '21-10-2019', '07-01-1973', '19-07-2014', '11-03-1992', '21-10-2019']
seen = set()
my_list = [datetime.datetime.strptime(d,'%d-%m-%Y')
for d in datelist
if d not in seen and not seen.add(d)]
my_list.sort(reverse=True)
You can use heapq.nlargest to do this.
import heapq
from datetime import datetime
datelist = [
'14-04-2001',
'29-12-2061',
'21-10-2019',
'07-01-1973',
'19-07-2014',
'11-03-1992',
'21-10-2019'
]
heapq.nlargest(3, {datetime.strptime(d, "%d-%m-%Y") for d in datelist})[-1]
This return datetime.datetime(2014, 7, 19, 0, 0)
I have a list of variable that includes several datetime.datetime type variables.
I.e.:
a['PrTimeStamp'] # Type list
[0] datetime.datetime(2014, 10, 19, 10, 0) # Type datetime
[1] datetime.datetime(2014, 12, 3, 12, 0) # Type datetime
[2] datetime.datetime(2014, 12, 4, 0, 0) # Type datetime
[3] datetime.datetime(2014, 12, 10, 13, 0) # Type datetime
[4] datetime.datetime(2014, 12, 16, 20, 0) # Type datetime
[5] datetime.datetime(2014, 12, 17, 2, 0) # Type datetime
E.g. if I make print(a['PrTimeStamp'][0]) it prints to screen: 2014-10-19 10:00:00.
Now I want to convert this list to a format that can be easily read by matlab.
I tried to do the following:
a['PrTimeStamp'] = a['PrTimeStamp'].strftime("%d-%b-%Y %H:%M:%S"))
But I got an error:
AttributeError: 'list' object has no attribute 'strftime'
date[]
for i in a['PrTimeStamp']:
date.append(i.strftime("%d-%b-%Y %H:%M:%S"))
a must be a dictionary has a key PrTimeStamp
I'm pretty sure you need data.strftime to format the dates. Something similar to this.
How to change time formats in python?
Since a['PrTimeStamp'] is a list of date time objects, you need to iterate each item in the list and format the date time object in it!
Since you haven't posted the entire code,
a={}
import datetime
a['PrTimeStamp'] = [ datetime.datetime(2014, 10, 19, 10, 0)
, datetime.datetime(2014, 12, 3, 12, 0)
, datetime.datetime(2014, 12, 4, 0, 0)
, datetime.datetime(2014, 12, 10, 13, 0)
, datetime.datetime(2014, 12, 16, 20, 0)
, datetime.datetime(2014, 12, 17, 2, 0)]
#this is what you need!
print [ts.strftime("%d-%b-%Y %H:%M:%S") for ts in a['PrTimeStamp']]
will print
['19-Oct-2014 10:00:00', '03-Dec-2014 12:00:00', '04-Dec-2014 00:00:00', '10-Dec-2014 13:00:00', '16-Dec-2014 20:00:00', '17-Dec-2014 02:00:00']
Hope it helps!
I have a column formatted as such in one of my models:
TEMP_START = models.DateTimeField(null=True)
And I am attempting to do an exact lookup using queryset syntax such as
x.filter(TEMP_START=my_datetime_object) # x can be thought of as y.objects.all()
This returns no objects, when it should do more than one. However,
x.filter(TEMP_START__date=my_datetime_object.date()).filter(TEMP_START__hour=my_datetime_object.hour)
Does return the proper objects (they're hourly). Are direct datetime filters not supported, and thus keywords must be used?
====== Edit with bad results:
Searching for: {'TEMP_START': datetime.datetime(2016, 3, 31, 2, 0)}
Values in column: [{'TEMP_START': datetime.datetime(2016, 3, 29, 8, 0)}, {'TEMP_START': datetime.datetime(2016, 3, 29, 14, 0)}, {'TEMP_START': datetime.datetime(2016, 3, 30, 2, 0)}, {'TEMP_START': datetime.datetime(2016, 3, 29, 20, 0)}, {'TEMP_START': datetime.datetime(2016, 3, 30, 8, 0)}, {'TEMP_START': datetime.datetime(2016, 3, 30, 20, 0)}, {'TEMP_START': datetime.datetime(2016, 3, 31, 2, 0)}, {'TEMP_START': datetime.datetime(2016, 3, 30, 14, 0)}]
Values being returned: []
Code:
args_timeframe_start = {variables.temp_start: self.ranked_timeframes[rank][variables.temp_start]}
print(args_timeframe_start)
print(self.query_data.values(variables.temp_start).distinct())
query_data = self.query_data.filter(**args_timeframe_start)
print(query_data.values(variables.temp_start).distinct())
You need to find out what is my_datetime_object but most likely because DateTime fields contain python datetime.datetime objects, datetime.datetime objects is composed of year, month, date, hour, minute, second, microsecond. So if you merely compare date and hour, sure you could get results, but you couldn't guarantee that my_datetime_object matches one of the records in your database that has the same minute, second and microsecond.
Try this quickly and you could see what does datetime look like, also django doc about DateTimeField:
from datetime import datetime
date = datetime.now()
print date