I've got a timedelta. I want the days, hour开发者_JAVA百科s and minutes from that - either as a tuple or a dictionary... I'm not fussed.
I must have done this a dozen times in a dozen languages over the years but Python usually has a simple answer to everything so I thought I'd ask here before busting out some nauseatingly simple (yet verbose) mathematics.
Mr Fooz raises a good point.
I'm dealing with "listings" (a bit like ebay listings) where each one has a duration. I'm trying to find the time left by doing when_added + duration - now
Am I right in saying that wouldn't account for DST? If not, what's the simplest way to add/subtract an hour?
If you have a datetime.timedelta
value td
, td.days
already gives you the "days" you want. timedelta
values keep fraction-of-day as seconds (not directly hours or minutes) so you'll indeed have to perform "nauseatingly simple mathematics", e.g.:
def days_hours_minutes(td):
return td.days, td.seconds//3600, (td.seconds//60)%60
This is a bit more compact, you get the hours, minutes and seconds in two lines.
days = td.days
hours, remainder = divmod(td.seconds, 3600)
minutes, seconds = divmod(remainder, 60)
# If you want to take into account fractions of a second
seconds += td.microseconds / 1e6
days, hours, minutes = td.days, td.seconds // 3600, td.seconds // 60 % 60
As for DST, I think the best thing is to convert both datetime
objects to seconds. This way the system calculates DST for you.
>>> m13 = datetime(2010, 3, 13, 8, 0, 0) # 2010 March 13 8:00 AM
>>> m14 = datetime(2010, 3, 14, 8, 0, 0) # DST starts on this day, in my time zone
>>> mktime(m14.timetuple()) - mktime(m13.timetuple()) # difference in seconds
82800.0
>>> _/3600 # convert to hours
23.0
For all coming along and searching for an implementation:
The above posts are related to datetime.timedelta, which is sadly not having properties for hours and seconds. So far it was not mentioned, that there is a package, which is having these. You can find it here:
- Check the source: https://github.com/andrewp-as-is/timedelta.py
- Available via pip: https://pypi.org/project/timedelta/
Example - Calculation:
>>> import timedelta
>>> td = timedelta.Timedelta(days=2, hours=2)
# init from datetime.timedelta
>>> td = timedelta.Timedelta(datetime1 - datetime2)
Example - Properties:
>>> td = timedelta.Timedelta(days=2, hours=2)
>>> td.total.seconds
180000
>>> td.total.minutes
3000
>>> td.total.hours
50
>>> td.total.days
2
I hope this could help someone...
I don't understand
days, hours, minutes = td.days, td.seconds // 3600, td.seconds // 60 % 60
how about this
days, hours, minutes = td.days, td.seconds // 3600, td.seconds % 3600 / 60.0
You get minutes and seconds of a minute as a float.
I used the following:
delta = timedelta()
totalMinute, second = divmod(delta.seconds, 60)
hour, minute = divmod(totalMinute, 60)
print(f"{hour}h{minute:02}m{second:02}s")
Here is a little function I put together to do this right down to microseconds:
def tdToDict(td:datetime.timedelta) -> dict:
def __t(t, n):
if t < n: return (t, 0)
v = t//n
return (t - (v * n), v)
(s, h) = __t(td.seconds, 3600)
(s, m) = __t(s, 60)
(micS, milS) = __t(td.microseconds, 1000)
return {
'days': td.days
,'hours': h
,'minutes': m
,'seconds': s
,'milliseconds': milS
,'microseconds': micS
}
Here is a version that returns a tuple
:
# usage: (_d, _h, _m, _s, _mils, _mics) = tdTuple(td)
def tdTuple(td:datetime.timedelta) -> tuple:
def _t(t, n):
if t < n: return (t, 0)
v = t//n
return (t - (v * n), v)
(s, h) = _t(td.seconds, 3600)
(s, m) = _t(s, 60)
(mics, mils) = _t(td.microseconds, 1000)
return (td.days, h, m, s, mics, mils)
While pandas.Timedelta
does not provide these attributes directly, it indeed provide a method called total_seconds
, based on which days, hours, and minutes can be easily derived:
import pandas as pd
td = pd.Timedelta("2 days 12:30:00")
minutes = td.total_seconds()/60
hours = minutes/60
days = hours/ 24
print(minutes, hours, days)
I found the easiest way is using str(timedelta)
. It will return a sting formatted like 3 days, 21:06:40.001000
, and you can parse hours and minutes using simple string operations or regular expression.
While if you are using python datetime package, you can also code like below:
import datetime
tap_in = datetime.datetime.strptime("04:12", "%H:%M")
tap_out = datetime.datetime.strptime("18:20", "%H:%M")
num_of_hour = (tap_out - tap_in).total_seconds()/3600
num_of_hour # 14.133333333333333
This is another possible approach, though a bit wordier than those already mentioned. It maybe isn't the best approach for this scenario but it is handy to be able to obtain your time duration in a specific unit that isn't stored within the object (weeks, hours, minutes, milliseconds) and without having to remember or calculate conversion factors.
from datetime import timedelta
one_hour = timedelta(hours=1)
one_minute = timedelta(minutes=1)
print(one_hour/one_minute) # Yields 60.0
I've got a timedelta. I want the days, hours and minutes from that - either as a tuple or a dictionary... I'm not fussed.
in_time_delta = timedelta(days=2, hours=18, minutes=30)
td_d = timedelta(days=1)
td_h = timedelta(hours=1)
td_m = timedelta(minutes=1)
dmh_list = [in_time_delta.days,
(in_time_delta%td_d)//td_h,
(in_time_delta%td_h)//td_m]
Which should assign [2, 18, 30]
to dmh_list
If using pandas (at least version >1.0), the Timedelta
class has a components
attribute that returns a named tuple with all the fields nicely laid out.
e.g.
import pandas as pd
delta = pd.Timestamp("today") - pd.Timestamp("2022-03-01")
print(delta.components)
timedelta = pd.Timestamp("today") - pd.Timestamp("2022-01-01")
print(timedelta.components)
print(timedelta.components.days)
print(timedelta.components.seconds)
will return something like:
Components(days=281, hours=2, minutes=24, seconds=3, milliseconds=72, microseconds=493, nanoseconds=0)
281
3
timedeltas have a days
and seconds
attribute .. you can convert them yourself with ease.
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