I have a function that returns information in seconds, but I need to store that information in hours:minutes:seconds.
Is there an easy way to convert the seconds to this format in Python?
Answers:
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Method 1
You can use datetime.timedelta function:
>>> import datetime >>> str(datetime.timedelta(seconds=666)) '0:11:06'
Method 2
By using the divmod() function, which does only a single division to produce both the quotient and the remainder, you can have the result very quickly with only two mathematical operations:
m, s = divmod(seconds, 60) h, m = divmod(m, 60)
And then use string formatting to convert the result into your desired output:
print('{:d}:{:02d}:{:02d}'.format(h, m, s)) # Python 3
print(f'{h:d}:{m:02d}:{s:02d}') # Python 3.6+
Method 3
I can hardly name that an easy way (at least I can’t remember the syntax), but it is possible to use time.strftime, which gives more control over formatting:
from time import strftime
from time import gmtime
strftime("%H:%M:%S", gmtime(666))
'00:11:06'
strftime("%H:%M:%S", gmtime(60*60*24))
'00:00:00'
gmtime is used to convert seconds to special tuple format that strftime() requires.
Note: Truncates after 23:59:59
Method 4
Using datetime:
With the ':0>8' format:
from datetime import timedelta
"{:0>8}".format(str(timedelta(seconds=66)))
# Result: '00:01:06'
"{:0>8}".format(str(timedelta(seconds=666777)))
# Result: '7 days, 17:12:57'
"{:0>8}".format(str(timedelta(seconds=60*60*49+109)))
# Result: '2 days, 1:01:49'
Without the ':0>8' format:
"{}".format(str(timedelta(seconds=66)))
# Result: '00:01:06'
"{}".format(str(timedelta(seconds=666777)))
# Result: '7 days, 17:12:57'
"{}".format(str(timedelta(seconds=60*60*49+109)))
# Result: '2 days, 1:01:49'
Using time:
from time import gmtime
from time import strftime
# NOTE: The following resets if it goes over 23:59:59!
strftime("%H:%M:%S", gmtime(125))
# Result: '00:02:05'
strftime("%H:%M:%S", gmtime(60*60*24-1))
# Result: '23:59:59'
strftime("%H:%M:%S", gmtime(60*60*24))
# Result: '00:00:00'
strftime("%H:%M:%S", gmtime(666777))
# Result: '17:12:57'
# Wrong
Method 5
This is my quick trick:
from humanfriendly import format_timespan
secondsPassed = 1302
format_timespan(secondsPassed)
# '21 minutes and 42 seconds'
For more info Visit:
https://humanfriendly.readthedocs.io/en/latest/#humanfriendly.format_timespan
Method 6
The following set worked for me.
def sec_to_hours(seconds):
a=str(seconds//3600)
b=str((seconds%3600)//60)
c=str((seconds%3600)%60)
d=["{} hours {} mins {} seconds".format(a, b, c)]
return d
print(sec_to_hours(10000))
# ['2 hours 46 mins 40 seconds']
print(sec_to_hours(60*60*24+105))
# ['24 hours 1 mins 45 seconds']
Method 7
This is how I got it.
def sec2time(sec, n_msec=3):
''' Convert seconds to 'D days, HH:MM:SS.FFF' '''
if hasattr(sec,'__len__'):
return [sec2time(s) for s in sec]
m, s = divmod(sec, 60)
h, m = divmod(m, 60)
d, h = divmod(h, 24)
if n_msec > 0:
pattern = '%%02d:%%02d:%%0%d.%df' % (n_msec+3, n_msec)
else:
pattern = r'%02d:%02d:%02d'
if d == 0:
return pattern % (h, m, s)
return ('%d days, ' + pattern) % (d, h, m, s)
Some examples:
$ sec2time(10, 3) Out: '00:00:10.000' $ sec2time(1234567.8910, 0) Out: '14 days, 06:56:07' $ sec2time(1234567.8910, 4) Out: '14 days, 06:56:07.8910' $ sec2time([12, 345678.9], 3) Out: ['00:00:12.000', '4 days, 00:01:18.900']
Method 8
If you need to get datetime.time value, you can use this trick:
my_time = (datetime(1970,1,1) + timedelta(seconds=my_seconds)).time()
You cannot add timedelta to time, but can add it to datetime.
UPD: Yet another variation of the same technique:
my_time = (datetime.fromordinal(1) + timedelta(seconds=my_seconds)).time()
Instead of 1 you can use any number greater than 0. Here we use the fact that datetime.fromordinal will always return datetime object with time component being zero.
Method 9
hours (h) calculated by floor division (by //) of seconds by 3600 (60 min/hr * 60 sec/min)
minutes (m) calculated by floor division of remaining seconds (remainder from hour calculation, by %) by 60 (60 sec/min)
similarly, seconds (s) by remainder of hour and minutes calculation.
Rest is just string formatting!
def hms(seconds):
h = seconds // 3600
m = seconds % 3600 // 60
s = seconds % 3600 % 60
return '{:02d}:{:02d}:{:02d}'.format(h, m, s)
print(hms(7500)) # Should print 02h05m00s
Method 10
A bit off topic answer but maybe useful to someone
def time_format(seconds: int):
if seconds is not None:
seconds = int(seconds)
d = seconds // (3600 * 24)
h = seconds // 3600 % 24
m = seconds % 3600 // 60
s = seconds % 3600 % 60
if d > 0:
return '{:02d}D {:02d}H {:02d}m {:02d}s'.format(d, h, m, s)
elif h > 0:
return '{:02d}H {:02d}m {:02d}s'.format(h, m, s)
elif m > 0:
return '{:02d}m {:02d}s'.format(m, s)
elif s > 0:
return '{:02d}s'.format(s)
return '-'
Results in:
print(time_format(25*60*60 + 125)) >>> 01D 01H 02m 05s print(time_format(17*60*60 + 35)) >>> 17H 00m 35s print(time_format(3500)) >>> 58m 20s print(time_format(21)) >>> 21s
Method 11
dateutil.relativedelta is convenient if you need to access hours, minutes and seconds as floats as well. datetime.timedelta does not provide a similar interface.
from dateutil.relativedelta import relativedelta
rt = relativedelta(seconds=5440)
print(rt.seconds)
print('{:02d}:{:02d}:{:02d}'.format(
int(rt.hours), int(rt.minutes), int(rt.seconds)))
Prints
40.0 01:30:40
Method 12
You can divide seconds by 60 to get the minutes
import time seconds = time.time() minutes = seconds / 60 print(minutes)
When you divide it by 60 again, you will get the hours
Method 13
In my case I wanted to achieve format
“HH:MM:SS.fff”.
I solved it like this:
timestamp = 28.97000002861023
str(datetime.fromtimestamp(timestamp)+timedelta(hours=-1)).split(' ')[1][:12]
'00:00:28.970'
Method 14
The solutions above will work if you’re looking to convert a single value for “seconds since midnight” on a date to a datetime object or a string with HH:MM:SS, but I landed on this page because I wanted to do this on a whole dataframe column in pandas. If anyone else is wondering how to do this for more than a single value at a time, what ended up working for me was:
mydate='2015-03-01'
df['datetime'] = datetime.datetime(mydate) +
pandas.to_timedelta(df['seconds_since_midnight'], 's')
Method 15
Here is a way that I always use: (no matter how inefficient it is)
seconds = 19346
def zeroes (num):
if num < 10: num = "0" + num
return num
def return_hms(second, apply_zeroes):
sec = second % 60
min_ = second // 60 % 60
hrs = second // 3600
if apply_zeroes > 0:
sec = zeroes(sec)
min_ = zeroes(min_)
if apply_zeroes > 1:
hrs = zeroes(hrs)
return "{}:{}:{}".format(hrs, min_, sec)
print(return_hms(seconds, 1))
RESULT:
5:22:26
Syntax of return_hms() function
The return_hms() function is used like this:
The first variable (second) is the amount of seconds you want to convert into h:m:s.
The second variable (apply_zeroes) is formatting:
0 or less: Apply no zeroes whatsoever
1: Apply zeroes to minutes and seconds when they’re below 10.
2 or more: Apply zeroes to any value (including hours) when they’re below 10.
Method 16
Here is a simple program that reads the current time and converts it to a time of day in hours, minutes, and seconds
import time as tm #import package time
timenow = tm.ctime() #fetch local time in string format
timeinhrs = timenow[11:19]
t=tm.time()#time.time() gives out time in seconds since epoch.
print("Time in HH:MM:SS format is: ",timeinhrs,"nTime since epoch is : ",t/(3600*24),"days")
The output is
Time in HH:MM:SS format is: 13:32:45 Time since epoch is : 18793.335252338384 days
Method 17
I looked every answers here and still tried my own
def a(t):
print(f"{int(t/3600)}H {int((t/60)%60) if t/3600>0 else int(t/60)}M {int(t%60)}S")
Results:
>>> a(7500) 2H 5M 0S >>> a(3666) 1H 1M 6S
Python: 3.8.8
Method 18
division = 3623 // 3600 #to hours
division2 = 600 // 60 #to minutes
print (division) #write hours
print (division2) #write minutes
PS My code is unprofessional
All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0