How to convert an H:MM:SS time string to seconds in Python?

Question:

Basically I have the inverse of this problem: Python Time Seconds to h:m:s

I have a string in the format H:MM:SS (always 2 digits for minutes and seconds), and I need the integer number of seconds that it represents. How can I do this in python?

For example:

  • “1:23:45” would produce an output of 5025
  • “0:04:15” would produce an output of 255
  • “0:00:25” would produce an output of 25

etc

Asked By: hughes

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Answers:

parts = time_string.split(":")
seconds = int(parts[0])*(60*60) + int(parts[1])*60 + int(parts[2])
Answered By: DaClown
def get_sec(time_str):
    """Get seconds from time."""
    h, m, s = time_str.split(':')
    return int(h) * 3600 + int(m) * 60 + int(s)


print(get_sec('1:23:45'))
print(get_sec('0:04:15'))
print(get_sec('0:00:25'))
Answered By: taskinoor
ts = '1:23:45'
secs = sum(int(x) * 60 ** i for i, x in enumerate(reversed(ts.split(':'))))
print(secs)
Answered By: FMc

Without many checks, and assuming it’s either "SS" or "MM:SS" or "HH:MM:SS" (although not necessarily two digits per part):

def to_seconds(timestr):
    seconds= 0
    for part in timestr.split(':'):
        seconds= seconds*60 + int(part, 10)
    return seconds

>>> to_seconds('09')
9
>>> to_seconds('2:09')
129
>>> to_seconds('1:02:09')
3729

This is a different “spelling” of FMc’s answer 🙂

Answered By: tzot

Using datetime module

import datetime
t = '10:15:30'
h,m,s = t.split(':')
print(int(datetime.timedelta(hours=int(h),minutes=int(m),seconds=int(s)).total_seconds()))

Output: 36930

Answered By: kaush

Another alternative if you have days on string:

def duration2sec(string):
    if "days" in string:
        days = string.split()[0]
        hours = string.split()[2].split(':')
        return int(days) * 86400 + int(hours[0]) * 3600 + int(hours[1]) * 60 + int(hours[2])
    else:
        hours = string.split(':')
        return int(hours[0]) * 3600 + int(hours[1]) * 60 + int(hours[2])
Answered By: Italux

You can use lambda and reduce a list and the fact that m=60s and h=60m. (see “Reducing a List” at http://www.python-course.eu/lambda.php)

timestamp = "1:23:45"
seconds = reduce(lambda x, y: x*60+y, [int(i) for i in (timestamp.replace(':',',')).split(',')])
Answered By: scavara

I didn’t really like any of the given answers, so I used the following:

def timestamp_to_seconds(t):
    return sum(float(n) * m for n,
               m in zip(reversed(time.split(':')), (1, 60, 3600))
               )
Answered By: Nathan Rice

You can split the time into a list and add each individual time component, multiplying the hours component by 3600 (the number of seconds in an hour) and the minutes component by 60 (number of seconds in a minute), like:

timeInterval ='00:35:01'
list = timeInterval.split(':')
hours = list[0]
minutes = list[1]
seconds = list[2]
total = (int(hours) * 3600 + int(minutes) * 60 + int(seconds))
print("total = ", total)
Answered By: raviGupta

Expanding on @FMc’s solution which embodies half of Horner’s method. Advantage of Horner’s method: Skip reversing the list, avoid power calculation.

from functools import reduce

timestamp = "1:23:45"
seconds = reduce(lambda s, d: s * 60 + int(d), timestamp.split(":"), 0)

or, if you dislike reduce (as does Guido van Rossum and @0xc0de below):

timestamp = "1:23:45"
seconds = 0
for d in timestamp.split(":"):
    seconds = seconds * 60 + int(d)

If you prefer zip (as does @nathan-rice below):

from itertools import accumulate, repeat
from operator import mul

def timestamp_to_seconds(t):
    return sum(int(n) * m for n, m in
       zip(reversed(t.split(":")), accumulate(repeat(60), func=mul, initial=1)))

Answered By: Bernhard Wagner

Using datetime module is also posible and more robust

import datetime as dt

def get_total_seconds(stringHMS):
   timedeltaObj = dt.datetime.strptime(stringHMS, "%H:%M:%S") - dt.datetime(1900,1,1)
   return timedeltaObj.total_seconds()

datetime.strptime parses the string according to format %H:%M:%S, and it creates a datetime object as year 1900, month1, day 1, hour H, minute M, and second S.

That’s why to get the total of seconds is necessary to subtract the year, month and day.

print(get_total_seconds('1:23:45'))
>>> 5025.0

print(get_total_seconds('0:04:15'))
>>> 255.0

print(get_total_seconds('0:00:25'))
>>>25.0
Answered By: DaniTeba

Just a simple generalization to the great response of taskinoor

In the context of my problem the format is similar, but includes AM or PM.

Format ‘HH:MM:SS AM’ or ‘HH:MM:SS PM’

For this case the function changes to:

def get_sec(time_str):
"""Get Seconds from time."""
if 'AM' in time_str:
    time_str = time_str.strip('AM')
    h, m, s = time_str.split(':')
    seconds = int(h) * 3600 + int(m) * 60 + int(s)
if 'PM' in time_str:
    time_str = time_str.strip('PM')
    h, m, s = time_str.split(':')
    seconds = (12 + int(h)) * 3600 + int(m) * 60 + int(s)

return seconds 
Answered By: Alejandro

In Pandas using @JayRizzo ‘s cool function and a list comprehension:

def get_sec(time_str):
    """Get Seconds from time."""
    h, m, s = time_str.split(':')
    return int(h) * 3600 + int(m) * 60 + int(s)

df['secs']=[get_sec(x) for x in df['original_time_string']]
Answered By: Manolianos

For %H:%M:%S.%f

def get_sec(time_str):
    h, m, s = time_str.split(':')
    return int(h) * 3600 + int(m) * 60 + float(s)
Answered By: Ivan_Aka

I liked tzot’s answer above the best for my purposes because my H:M:S are human created and the can have float seconds part
Here is my small modification

def hms_to_seconds(timestr:str)->float:
    """Get seconds from time.

    :param timestr: hh:mm:ss.xxx string or mm:s.xxx or simply s.xxx where xxx is the fraction of seconds
    :returns: time in float seconds
    """
    seconds= 0.
    for part in timestr.split(':'):
        seconds= seconds*60. + float(part)
    return seconds

>>> hms_to_seconds('3.00')
3.0
>>> hms_to_seconds('1:3.00')
63.0
>>> hms_to_seconds('1:0:3.5')
3603.5

Answered By: tobi delbruck
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