Converting a float to hh:mm format

Question

I want to print a float value that I've got after performing some math as time string in the hh:mm format. Currently I have the float like 9.888888888888886 and I want it in time like 09:50. I have tried the following code:

    time = str(time)
    time = time.split(".")
    time[1] = float(time[1])
    time[1] *= 0.6
    time[1] = str(time[1])

and when I print I use

    str(time[0]) + ":" + time[1][:2]

Any way to achieve this effect consistently? With more advanced inputs my above code does not work properly, and outputs the wrong time.

Define "does not work properly". Define "wrong time". – Sam Varshavchik Dec 16 '14 at 03:10 — Sam Varshavchik, Dec 16 '14 at 03:10

score 18 · Accepted Answer · edited Feb 15 '21 at 11:13

18

Alternatively to using datetime, you can just use simple formatting with a bit of maths, eg:

result = '{0:02.0f}:{1:02.0f}'.format(*divmod(time * 60, 60))

(This assumes that time is nonnegative.)

edited Feb 15 '21 at 11:13

Community

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answered Dec 16 '14 at 03:17

Jon Clements

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what if you want to have it in the format HH:MM:SS? – FaCoffee Sep 12 '18 at 11:43
2

It does not round correctly, `time=0.999` becomes `00:60` instead of `01:00` – Adam May 26 '21 at 10:05

falsetru · Answer 2 · 2014-12-16T04:06:31.770

Using datetime.timedelta:

>>> import datetime
>>> datetime.timedelta(hours=9.888888888888886)
datetime.timedelta(0, 35600)
>>> str(datetime.timedelta(hours=9.888888888888886))
'9:53:20'
>>> str(datetime.timedelta(hours=9.888888888888886)).rsplit(':', 1)[0]
'9:53'

>>> datetime.datetime.combine(datetime.date.today(), datetime.time()) + \
... datetime.timedelta(hours=9.888888888888886)
datetime.datetime(2014, 12, 16, 9, 53, 20)
>>> _.strftime('%H:%M')
'09:53'

score 9 · Answer 3 · answered Dec 16 '14 at 03:29

Using divmod you get the remainder and the quotient of a division.

minutes = 9.888888888888886*60
hours, minutes = divmod(minutes, 60)

print "%02d:%02d"%(hours,minutes)

Output:

09:53

If you want the seconds too:

seconds = 9.888888888888886*60*60
minutes, seconds = divmod(seconds, 60)
hours, minutes = divmod(minutes, 60)

print "%02d:%02d:%02d"%(hours,minutes,seconds)

ZF007 · Answer 4 · 2019-07-13T11:23:12.883

For python 3.7 version this becomes an one-liner.

x = 9.888888888888886

print (str(datetime.timedelta(hours=x))[:-3])

(Here the time-string minus the last three chars is printed.)

Result 1 : 9:53

Alternatively if you needs seconds:

print (datetime.timedelta(hours=x))

Result 2 : 9:53:20

And finally if you go beyond the 24hour mark timedelta shows the added day as well:

x = 39.888888888888886

Result 3 : 1 day, 15:53:20

Gagravarr · Answer 5 · 2014-12-16T03:26:22.223

Assuming you want to turn 9.888888888888886 into 09:53, based on the former being floating point hours, and you don't want to worry about days, without a library you could simply do:

fhours = 9.888888888888886
ihours = int(ihours)
"%02d:%02d" % (ihours,(fhours-ihours)*60)

With only the math module, try

import math
"%02d:%02d" % (fhours,math.modf(fhours)[0]*60)

You might think that using the datetime module would help, but it's actually a bit fiddly. You can get a timedelta object out by passing in your fractional hours:

 import datetime
 td = datetime.timedelta(hours=9.888888888888886)

But getting that into a time object or a datetime object to format is the tricky bit, you end up having to do something like

 import datetime
 td = datetime.timedelta(hours=9.888888888888886)
 (datetime.datetime(2000,1,1)+td).strftime("%H:%M")

(That uses a randomly picked date to get the time delta into a datetime to use)

Converting a float to hh:mm format

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