This question already has answers here:
Stripping off the seconds in datetime python
(5 answers)
Closed 4 years ago.
When i preform datetime.now() to get the current time, it gets printed (with print) like this:
2018-06-14 13:21:02.331933
I want to ignore seconds and microseconds and to have it printed out like this:
2018-06-14 13:21:00
I know that I can convert to a string of my choosing with strftime but i'm not looking to alter the string but to change the datetime object itself so that it has 0 seconds and 0 milliseconds (and it seems stupid to do strftime and immediately strptime to convert back).
In other words to convert from datetime(2018, 6, 14, 14, 7, 27, 326853) to datetime(2018, 6, 14, 14, 7)
I've tried like this:
now = datetime.now()
now = now - timedelta(seconds=now.second, microseconds=now.microsecond)
And that works, but i'm sure there's a simpler and more efficient way (I need to do this thousands of times when iterating through different files). So my question is if i'm right that there is a simpler way and if so what are your solutions to such a problem?
Use datetime.replace:
now = datetime.now().replace(second=0, microsecond=0)
Use the datetime.replace() method to create a new instance, with specific attributes set to your desired value:
now = datetime.now().replace(second=0, microsecond=0)
Demo:
>>> from datetime import datetime
>>> datetime.now()
datetime.datetime(2018, 6, 14, 12, 16, 4, 738362)
>>> datetime.now().replace(second=0, microsecond=0)
datetime.datetime(2018, 6, 14, 12, 16)
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 am subtracting two dates in python 2.7 and getting wrong result in seconds. Apparently difference between these dates is more than 24h which is 86400s. But I am getting 44705s, why is that and how to fix it ?
>>> date1
datetime.datetime(2017, 10, 22, 11, 41, 28)
>>> date2
datetime.datetime(2017, 10, 20, 23, 16, 23)
>>> (date1-date2).seconds
44705
Calling .seconds will only give you the seconds component of the timedelta object, which only takes into account seconds, minutes, and hours (see docs for more detail). If you want the entire timedelta in seconds, call total_seconds.
>>> (date1 - date2).total_seconds()
131105.0
date1-date2 is datetime.timedelta(1, 44705). You're only looking at the seconds portion. Look at the days portion too.
I have a unix timestamp in seconds (such as 1294778181) that I can convert to UTC using
from datetime import datetime
datetime.utcfromtimestamp(unix_timestamp)
Problem is, I would like to get the corresponding time in 'US/Eastern' (considering any DST) and I cannot use pytz and other utilities.
Only datetime is available to me.
Is that possible?
Thanks!
Easiest, but not supersmart solution is using timedelta
import datetime
>>> now = datetime.datetime.utcnow()
US/Eastern is 5 hours behind UTC, so let's just create thouse five hours as a timedelta object and make it negative, so that when reading back our code we can see that the offset is -5 and that there's no magic to deciding when to add and when to subtract timezone offset
>>> eastern_offset = -(datetime.timedelta(hours=5))
>>> eastern = now + eastern_offset
>>> now
datetime.datetime(2016, 8, 26, 20, 7, 12, 375841)
>>> eastern
datetime.datetime(2016, 8, 26, 15, 7, 12, 375841)
If we wanted to fix DST, we'd run the datetime through smoething like this (not entirely accurate, timezones are not my expertise (googling a bit now it changes each year, yuck))
if now.month > 2 and now.month < 12:
if (now.month == 3 and now.day > 12) or (now.month == 11 and now.day < 5):
eastern.offset(datetime.timedelta(hours=5))
You could go even into more detail, add hours, find out how exactly it changes each year... I'm not going to go through all that :)
This question already has answers here:
How can I account for period (AM/PM) using strftime?
(5 answers)
Closed 7 years ago.
I'm trying to convert a string date to epoch, but it doesn't seem to pick up the am or pm. The time always defaults to am.
I've tried this:
dt = '2015-05-04 5:55PM'
pattern = '%Y-%m-%d %H:%M%p'
epochDate = int(time.mktime(time.strptime(dt, pattern)))
print epochDate
# Result
1430684700
# Checking output
datetime.datetime.fromtimestamp(1430684700).strftime('%Y-%m-%d %H:%M:%S%p')
# Doesn't show PM
'2015-05-04 05:55:00AM'
I'm not sure what I've done wrong here?
You should use the python-dateutil library:
>>> import dateutil.parser
>>> dateutil.parser.parse('2015-05-04 5:55PM')
datetime.datetime(2015, 5, 4, 17, 55)
>>> dateutil.parser.parse('2015-05-04 5:55AM')
datetime.datetime(2015, 5, 4, 5, 55)
I have time defined in the following format: "%A %H:%M:%S" i.e "Monday 06:00:00"
Now, using the above format, I define a time window:
"Monday 06:00:00" - "Monday 18:00:00"
I now want to check if the current time falls within the time window or not. To do that, I am using the datetime utility.
import datetime
# current time in str
current_time = datetime.datetime.strftime(datetime.datetime.now(),'%A %H:%M:%S')
# convert to datetime object
current_time = datetime.datetime.strptime(current_time, '%A %H:%M:%S')
# print current_time produces: 1900-01-01 19:13:53
# and day of week information is lost
However, the moment, I convert the current_time to a datetime object, the day of the week information is lost.
What's the best way to compare time windows that includes day-of-week in Python?
BTW: I want the time window to be repeated in the future. For eg, "Monday 06:00 - Tuesday 05:00" would apply to all weeks in the future.
datetime would know nothing about which monday you're talking about. There have been 1000's since the epoch.
I suggest you look into the dateutil package. It applies "human-ish" semantics:
>>> import dateutil.parser
>>> dateutil.parser.parse('Monday')
datetime.datetime(2014, 2, 3, 0, 0)
>>> dateutil.parser.parse('Monday 18:30:00')
datetime.datetime(2014, 2, 3, 18, 30)
Note how it assumes that Monday means today (it's Monday). And Tuesday and Sunday (below) mean this coming Tuesday, Sunday:
>>> dateutil.parser.parse('Tuesday 18:30:00')
datetime.datetime(2014, 2, 4, 18, 30)
>>> dateutil.parser.parse('Sunday 18:30:00')
datetime.datetime(2014, 2, 9, 18, 30)
To take this further one take a look at the rrule module/class in dateutil. And the built-in calendar module. For example, with rrule I can do:
next_five = rrule.rrule(
rrule.WEEKLY,
count=5,
dtstart=dateutil.parser.parse('Sunday 18:30:00'),
)
This returns an iterator of the next 5 weeks starting at 6:30pm Sunday. The iterator will produce
[datetime.datetime(2014, 2, 5, 18, 30),
datetime.datetime(2014, 2, 12, 18, 30),
datetime.datetime(2014, 2, 19, 18, 30),
datetime.datetime(2014, 2, 26, 18, 30),
datetime.datetime(2014, 3, 5, 18, 30)]
This should get you going on the final solution.
In my experience Datetime objects are YYYY:MM:DD HH:MM:SS (more or less). The day of week can be derived from that object using datetime's weekday() - see http://docs.python.org/2/library/datetime.html#datetime.date.weekday
You can solve your problem using greater/less than "<" and ">" operators for datetime evaluation. Here is a proof for simple comparison http://pastebin.com/miTUW9nF
Your logic would be (paraphrased): "if Xtime > windowstart and Xtime < windowend then Xtime is in window"
Datetime compares down to the second, my example is now and (now-10 days), that should cover your needs.