Reputation: 417
I need to convert the following strings into either datetimes or strings in a different format. My input looks like these examples:
196h 26m 13s
95h 19m
45m 28s
My desired string output would be (But my final goal is to convert these strings into datetime
objects):
196:26:13
95:19:00
00:45:28
In Excel, it would be [h]:mm:ss
.
Note: As you can see, hours can be higher than 24.
I have tried to parse this string with the time.strptime()
method, but it doesn't work with hours higher than 24. I have a solution with regular expressions, but I want to know if there is a more straightforward way to do this. What is the best way to solve this?
Upvotes: 1
Views: 312
Reputation: 85432
This would give you time deltas:
from datetime import timedelta
def to_timedelta(time_string):
units = {'h': 'hours', 'd': 'days', 'm': 'minutes', 's': 'seconds'}
return timedelta(**{units[x[-1]]: int(x[:-1]) for x in time_string.split()})
Test:
times = ['196h 26m 13s', '95h 19m', '45m 28s']
for t in times:
print(to_timedelta(t))
Output:
8 days, 4:26:13
3 days, 23:19:00
0:45:28
timedelta takes these arguments;
datetime.timedelta(days=0, seconds=0, microseconds=0, milliseconds=0, minutes=0, hours=0, weeks=0)
Using this mapping:
units = {'h': 'hours', 'd': 'days', 'm': 'minutes', 's': 'seconds'}
allows the short units in the string to be mapped to the corresponding names of the arguments.
Using Pythons **
syntax, the resulting dictionary can be used as single argument that will be converted in to the matching keyword arguments.
Upvotes: 2
Reputation: 243907
The first thing we should do is use a regular expression and use timedelta instead of datetime.
import datetime
import re
regex = re.compile(r'((?P<hours>\d+?)h)?((?P<minutes>\d+?)m)?((?P<seconds>\d+?)s)?')
def parse_time(time_str):
parts = regex.match(time_str)
if not parts:
return
parts = parts.groupdict()
time_params = {}
for (name, param) in parts.items():
if param:
time_params[name] = int(param)
return datetime.timedelta(**time_params)
L = ["196h 26m 13s", "95h 19m", "45m 28s"]
for l in L:
print(parse_time(l))
Output:
8 days, 4:00:00
3 days, 23:00:00
0:45:00
Upvotes: 1