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

possible duplicate of [Convert time string expressed as \[m|h|d|s|w\] to seconds in Python](http://stackoverflow.com/questions/3096860/convert-time-string-expressed-as-numbermhdsw-to-seconds-in-python) — , Jun 19 '11 at 14:10

score 134 · Accepted Answer · edited Mar 01 '22 at 21:57

134

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'))

edited Mar 01 '22 at 21:57

LeopardShark

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answered Jun 19 '11 at 14:11

taskinoor

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It's very nice and clear and compact but fails if the time does not have all the components, e.g. '3' or "1:21", or if the seconds part is a float. For these cases maybe the tzot solution below could be better fit, using float values – tobi delbruck Feb 19 '23 at 14:57

FMc · Answer 2 · 2020-07-05T01:06:12.427

66

ts = '1:23:45'
secs = sum(int(x) * 60 ** i for i, x in enumerate(reversed(ts.split(':'))))
print(secs)

edited Jul 05 '20 at 01:06

answered Jun 19 '11 at 14:22

FMc

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That's a really interesting technique. Thanks for sharing this. – hughes Jun 19 '11 at 14:24
What does that double-star operator do? – hughes Mar 11 '14 at 03:05
@hughes Exponentiation. – FMc Mar 11 '14 at 03:23
4

This superior code will also handle `s` and `m:s` strings like "53" and "2:41" correctly – qubodup Sep 20 '16 at 10:32
2

This is great. Change `int(x)` to `float(x)` and it'll handle decimal seconds, plus it doesn't need a leading zero for hour. VTT timestamps omit the leading 0 hour until hour > 1. – Marcos Sep 28 '22 at 00:31

kaush · Answer 3 · 2019-03-22T03:37:02.510

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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

edited Mar 22 '19 at 03:37

answered May 14 '14 at 22:52

kaush

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What does the output from your code look like given the OP's example input? – rbatt May 14 '14 at 23:21
You don't need `re` for this. Strings have a split method: `str.split()`. Much cleaner to just use `t.split(':')` instead of `re.split(':', t)`. – reynoldsnlp Oct 18 '18 at 01:52
I ended up using this solution, as it seemed to be the most standard way of doing it. I slightly altered the conversion of h,m,s to int using a comprehension list to make the conversion line cleaner, which ends up looking like this: `h, m, s = [int(x) for x in t.split(':')]`. – Prevok Feb 19 '21 at 22:39

tzot · Answer 4 · 2021-09-06T10:07:28.420

13

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 :)

edited Sep 06 '21 at 10:07

answered Jun 19 '11 at 14:46

tzot

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Really useful. Thanks. I like that you take into account for shorter strings that only contain seconds or minutes... – petezurich Sep 20 '20 at 11:27
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This is the best solution here. How come it's so under-voted!? It considers strings with missing `minutes` or `hours` part in a neat, single function, and without using any extra library. Kudos! – shivams May 22 '21 at 15:51

score 6 · Answer 5 · answered May 19 '20 at 11:52

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

This worked for me. I used "%H:%M:%S.%f" instead to handle decimal seconds — Tony M, Jan 27 '21 at 10:17

score 2 · Answer 6 · answered Apr 08 '17 at 15:53

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(',')])

score 2 · Answer 7 · edited Aug 24 '18 at 10:13

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)

score 2 · Answer 8 · answered Sep 09 '21 at 09:40

2

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 Sep 09 '21 at 09:40

Ivan_Aka

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score 1 · Answer 9 · edited Jul 13 '19 at 04:31

1

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))
               )

edited Jul 13 '19 at 04:31

JayRizzo

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answered Aug 30 '17 at 23:18

Nathan Rice

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Bernhard Wagner · Answer 10 · 2022-01-26T10:39:47.403

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)))

Goes a step forward in performance, but maybe a step backwards in readability, unless all your co-workers/fellow code readers are good with `reduced` readability. +1 for those cases, I am not a fan of reduce() though and would prefer to avoid these. — 0xc0de, Jan 24 '22 at 18:43

score 1 · Answer 11 · answered Jun 19 '11 at 14:10

1

parts = time_string.split(":")
seconds = int(parts[0])*(60*60) + int(parts[1])*60 + int(parts[2])

answered Jun 19 '11 at 14:10

DaClown

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Almost worked, I had to use int(...) as taskinoor suggested for it to work properly. – hughes Jun 19 '11 at 14:12

score 0 · Answer 12 · answered Oct 15 '14 at 23:47

0

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 Oct 15 '14 at 23:47

Italux

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The question asks for converting H:MM:SS time string to seconds not to convert days to to secs. Please have try to reword your answer for the question – Ram Oct 16 '14 at 00:15
The example is functional, `else` part will do that. But I agree that the simpler the better. – kukido Oct 16 '14 at 00:29

score 0 · Answer 13 · answered Nov 16 '20 at 20:18

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

score 0 · Answer 14 · edited Jun 22 '21 at 23:18

0

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']]

edited Jun 22 '21 at 23:18

Sven Eberth

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answered Jun 22 '21 at 18:38

Manolianos

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score 0 · Answer 15 · answered Feb 19 '23 at 15:05

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

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

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