Convert string to time format more than 24 hours format ( [h]:mm:ss ) without incrementing day-CodePudding

I need to convert string to time in hours format. I tried using the below But I didn't get the results that I expect.

inp_string = "70:30:00"
print(datetime.strptime(inp_string, "%H:%M:%S").time())

For this I'm getting the following error:

ValueError: time data '70:30:00' does not match format '%H:%M:%S'

If I enter the value of hours below 24 I'm getting good results;

new = '23:30:00'
print("value:",datetime.strptime(new, "%H:%M:%S").time())
print(type(datetime.strptime(new, "%H:%M:%S").time()))

output

value: 23:30:00
<class 'datetime.time'>

Like this, I have to convert a string into time format where the hours are greater than 24.

CodePudding user response：

You need to write your own parser to convert your str to timedelta (or find a library that does this for you).

Assuming your duration is always like HH:MM:SS, you can use something like this:

import datetime

def parse_duration(s):
    return list(map(lambda x: int(x), s.split(":")))

inp1 = "70:30:00"
[h1, m1, s1] = parse_duration(inp1)
d1 = datetime.timedelta(hours=h1, minutes=m1, seconds=s1)


inp2 = "00:30:00"
[h2, m2, s2] = parse_duration(inp2)
d2 = datetime.timedelta(hours=h2, minutes=m2, seconds=s2)

res = d1   d2

# Format res however you like

CodePudding user response：

inp_string = "70:30:00"
new_hour=(int(inp_string[0:2]))$
new_string=str(new_hour) (inp_string[2:8])
print(new_string)
print(datetime.strptime(new_string, "%H:%M:%S").time())