I\'m having trouble formatting a datetime.timedelta
object.
Here\'s what I\'m trying to do: I have a list of objects and one of the members of the cl
One liner. Since timedeltas do not offer datetime's strftime, bring the timedelta back to a datetime, and use stftime.
This can not only achieve the OP's requested format Hours:Minutes, now you can leverage the full formatting power of datetime's strftime, should your requirements change to another representation.
import datetime
td = datetime.timedelta(hours=2, minutes=10, seconds=5)
print(td)
print(datetime.datetime.strftime(datetime.datetime.strptime(str(td), "%H:%M:%S"), "%H:%M"))
Output:
2:10:05
02:10
This also solves the annoyance that timedeltas are formatted into strings as H:MM:SS rather than HH:MM:SS, which lead me to this problem, and the solution I've shared.
I had a similar problem with the output of overtime calculation at work. The value should always show up in HH:MM, even when it is greater than one day and the value can get negative. I combined some of the shown solutions and maybe someone else find this solution useful. I realized that if the timedelta value is negative most of the shown solutions with the divmod method doesn't work out of the box:
def td2HHMMstr(td):
'''Convert timedelta objects to a HH:MM string with (+/-) sign'''
if td < datetime.timedelta(seconds=0):
sign='-'
td = -td
else:
sign = ''
tdhours, rem = divmod(td.total_seconds(), 3600)
tdminutes, rem = divmod(rem, 60)
tdstr = '{}{:}:{:02d}'.format(sign, int(tdhours), int(tdminutes))
return tdstr
timedelta to HH:MM string:
td2HHMMstr(datetime.timedelta(hours=1, minutes=45))
'1:54'
td2HHMMstr(datetime.timedelta(days=2, hours=3, minutes=2))
'51:02'
td2HHMMstr(datetime.timedelta(hours=-3, minutes=-2))
'-3:02'
td2HHMMstr(datetime.timedelta(days=-35, hours=-3, minutes=-2))
'-843:02'
def td_format(td_object):
seconds = int(td_object.total_seconds())
periods = [
('year', 60*60*24*365),
('month', 60*60*24*30),
('day', 60*60*24),
('hour', 60*60),
('minute', 60),
('second', 1)
]
strings=[]
for period_name, period_seconds in periods:
if seconds > period_seconds:
period_value , seconds = divmod(seconds, period_seconds)
has_s = 's' if period_value > 1 else ''
strings.append("%s %s%s" % (period_value, period_name, has_s))
return ", ".join(strings)
A straight forward template filter for this problem. The built-in function int() never rounds up. F-Strings (i.e. f'') require python 3.6.
@app_template_filter()
def diffTime(end, start):
diff = (end - start).total_seconds()
d = int(diff / 86400)
h = int((diff - (d * 86400)) / 3600)
m = int((diff - (d * 86400 + h * 3600)) / 60)
s = int((diff - (d * 86400 + h * 3600 + m *60)))
if d > 0:
fdiff = f'{d}d {h}h {m}m {s}s'
elif h > 0:
fdiff = f'{h}h {m}m {s}s'
elif m > 0:
fdiff = f'{m}m {s}s'
else:
fdiff = f'{s}s'
return fdiff