Python: checking input time

Question

I have a problem, Im supposed to get user input in the form 10:10:10 (hh: mm: ss) or 10:10 (mm: ss) or 10 (ss). Now I need to check the welding parameters:

• If I get only seconds, there is no limit.
• If I get mm:ss then the seconds are limited to 0..59 and the minutes are unlimited.
• If I get hh:mm:ss , then both seconds and minutes are limited to 0..59, while the hours are not limited.

Then return the TimeDelta object.

The naive way is to write multiply if to test all of this. But they need a smoother path.

 val = "11:66:11" try: val = map(int, val.split(':')) except ValueError: return False if len(val) == 1: return val if len(val) == 2: if val[1] > 59: print "Bad seconds" return False if len(val) == 3: if val[2] > 59 or val[1] >59: print "Bad seconds / minutes" return False while len(val) < 3: split.insert(0,0) return = timedelta(hours=split[0],minutes=split[1],seconds=split[2]) 

Answer 1

How about using regex here:

 import re import datetime pattern = re.compile(r'^(\d+)(?::([0-5]?\d)(?::([0-5]?\d))?)?$') def str2seconds(val): match = pattern.match(val) if not match: raise ValueError("Invalid input: %s" % val) else: result = 0 for i in match.groups(): if i is not None: result *= 60 result += int(i) return datetime.timedelta(seconds=result) 

Example:

 >>> print(str2seconds('255')) 0:04:15 >>> print(str2seconds('255:25')) 4:15:25 >>> print(str2seconds('255:25:25')) 10 days, 15:25:25 >>> print(str2seconds('255:25:25:255')) Traceback (most recent call last): File "<stdin>", line 1, in <module> File "y.py", line 8, in str2seconds raise ValueError("Invalid input: %s" % val) ValueError: Invalid input: 255:25:25:255 >>> print(str2seconds('255:25:60')) Traceback (most recent call last): File "<stdin>", line 1, in <module> File "y.py", line 8, in str2seconds raise ValueError("Invalid input: %s" % val) ValueError: Invalid input: 255:25:60 

---

Part of regex in parts:

• ^ : start of line
• (\d+) : 1st digit recorded as group 1
• (?::([0-5]?\d)(?::([0-5]?\d))?)? additional part:
  • (?:...) - not an exciting group
  • : matches literal :
  • ([0-5]?\d) optional digit 0-5, followed by the digit recorded as group 2
  • there is no capture in the first group: (?::([0-5]?\d))? which does not necessarily coincide with the second : followed by an optional digit 0-5, followed by a digit; and capturing numbers in group 3
  • and finally $ matches the end of the line. Strictly, ^ is not required at the beginning, since match binds a match at the beginning; however, $ is required, because otherwise the match will not be fixed at the end of the line! (Python 3.4 added re.fullmatch to fix this.)

match.groups() will be a tuple of 3 elements; mismatch groups will be returned as None .