CODEWITHSUNDEEP
pythondatetimetimedelta
user8864523
user8864523

Reputation:

Converting time to a float

I wrote a program to compute the hours I've worked but at the end, it returns the value as a time.

EX: I worked from 11:45 to 4:30. My program returns 4:45 I would like it to return 4.75.

How do I convert this?

I am using the following code:

import datetime as dt

from datetime import timedelta

start =(input("Enter start time: " ))
end =(input("Enter start time: " ))
start_dt = dt.datetime.strptime(start, '%H:%M')
end_dt = dt.datetime.strptime(end, '%H:%M')
time =(end_dt - start_dt)
if time.days < 0:
    time=timedelta(days=0,seconds=time.seconds,microseconds=time.microseconds)
    time= str(time)
    time2=dt.datetime.strptime(time, '%H:%M:%S') 
    off_set=dt.datetime.strptime('12:00:00', '%H:%M:%S')
    time= time2 - off_set 

print (time)

Upvotes: 6

Views: 45777

Answers (2)

Karel Marik
Karel Marik

Reputation: 1003

If you use datetime then I suggest:

>> import datetime
>> a = datetime.datetime.now()
datetime.datetime(2020, 5, 16, 11, 13, 59, 543985)
>> (a - datetime.datetime.combine(a.date(), datetime.time())).total_seconds() / 3600
11.233206662499999

Upvotes: 2

WNG
WNG

Reputation: 3805

If you use a datetime, the hour and minute method should give you what you want

>>>a=datetime.datetime.now().time()
>>>a
datetime.time(21, 17, 35, 562000) 
>>>a.hour+a.minute/60.0
21.283333333333335

If you are using a timedelta (as is the case in your code), you should use :

mytimedelta = atime - anothertime
secs=mytimedelta.seconds #timedelta has everything below the day level stored in seconds
minutes = ((secs/60)%60)/60.0
hours = secs/3600
print hours + minutes

Upvotes: 8

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