Python summing up time

Question

In python how do I sum up the following time?

 0:00:00
 0:00:15
 9:30:56

Answer 1

Bellow is a solution using list comprehension:

from datetime import timedelta

def time_sum(time: List[str]) -> timedelta:
    """
    Calculates time from list of time hh:mm:ss format
    """

    return sum(
        [
            timedelta(hours=int(ms[0]), minutes=int(ms[1]), seconds=int(ms[2]))
            for t in time
            for ms in [t.split(":")]
        ],
        timedelta(),
    )

Example:

time_list = ["0:00:00", "0:00:15", "9:30:56"]
total = time_sum(time_list)
print(f"Total time: {total}")

Answer 2

It depends on the form you have these times in, for example if you already have them as datetime.timedeltas, then you could just sum them up:

>>> s = datetime.timedelta(seconds=0) + datetime.timedelta(seconds=15) + datetime.timedelta(hours=9, minutes=30, seconds=56)
>>> str(s)
'9:31:11'

Answer 3

I'm really disappointed if there is not any more pythonic solution... :(

Horrible one ->

timeList = [ '0:00:00', '0:00:15', '9:30:56' ]

ttt = [map(int,i.split()[-1].split(':')) for i in timeList]
seconds=reduce(lambda x,y:x+y[0]*3600+y[1]*60+y[2],ttt,0)
#seconds == 34271

This one looks horrible too ->

zero_time = datetime.datetime.strptime('0:0:0', '%H:%M:%S')
ttt=[datetime.datetime.strptime(i, '%H:%M:%S')-zero_time for i in timeList]
delta=sum(ttt,zero_time)-zero_time
# delta==datetime.timedelta(0, 34271)

# str(delta)=='9:31:11' # this seems good, but 
# if we have more than 1 day we get for example str(delta)=='1 day, 1:05:22'

Really frustrating is also this ->

sum(ttt,zero_time).strftime('%H:%M:%S')  # it is only "modulo" 24 :( 

I really like to see one-liner so, I tried to make one in python3 :P (good result but horrible look)

import functools
timeList = ['0:00:00','0:00:15','9:30:56','21:00:00'] # notice additional 21 hours!
sum_fnc=lambda ttt:(lambda a:'%02d:%02d:%02d' % (divmod(divmod(a,60)[0],60)+(divmod(a,60)[1],)))((lambda a:functools.reduce(lambda x,y:x+y[0]*3600+y[1]*60+y[2],a,0))((lambda a:[list(map(int,i.split()[-1].split(':'))) for i in a])(ttt)))
# sum_fnc(timeList) -> '30:40:11'

Answer 4

As a list of strings?

timeList = [ '0:00:00', '0:00:15', '9:30:56' ]
totalSecs = 0
for tm in timeList:
    timeParts = [int(s) for s in tm.split(':')]
    totalSecs += (timeParts[0] * 60 + timeParts[1]) * 60 + timeParts[2]
totalSecs, sec = divmod(totalSecs, 60)
hr, min = divmod(totalSecs, 60)
print "%d:%02d:%02d" % (hr, min, sec)

Result:

9:31:11

Answer 5

Using timedeltas (tested in Python 3.4):

import datetime

timeList = ['0:00:00', '0:00:15', '9:30:56']
sum = datetime.timedelta()
for i in timeList:
    (h, m, s) = i.split(':')
    d = datetime.timedelta(hours=int(h), minutes=int(m), seconds=int(s))
    sum += d
print(str(sum))

Result:

9:31:11

Answer 6

Assuming you want to add up the seconds for a total time:

def parse_time(s):
    hour, min, sec = s.split(':')
    try:
        hour = int(hour)
        min = int(min)
        sec = int(sec)
    except ValueError:
        # handle errors here, but this isn't a bad default to ignore errors
        return 0
    return hour * 60 * 60 + min * 60 + sec

print parse_time('0:00:00') + parse_time('0:00:15') + parse_time('9:30:56')