Time-Calculator Solution Review

Python
1

Please let me know how I can improve my code, or maybe a different way of solving it. There are a few lines that could be expanded into multiple lines for better readability, but it’s still pretty clean. Enjoy!

Here is my solution to the Time-Calculator Exercise:

def add_time(start,add,day=None):
  time, period = start.split() 
  hour, minutes = map(int,time.split(':')) # converting everything to int 
  addH, addM = map(int,add.split(':'))
  midday = ('PM','AM') 
  new_day = '' 
  later = ''
  
  # Clockwork: 
  # These calculations essentially do the math for capping minutes to 59 and hours to 12
  # Divmod and tuple assigment!
  carry, minutes = divmod(minutes + addM,60) # 'carry' is 1 (minutes >= 60) or 0 (minutes <= 59)
  hour += carry 
  cycles, hour = divmod(hour + addH,12) # 'cycles' is # of 12-hours (half days) that 'hour' exceeds 
  
  # AM or PM:
  # We can also think of 'cycles' as the # of switches between AM and PM, this means that an even number of switches doesn't change the period but an odd number will
  # Use this logic to subtract 1(odd cycles) or 0(even cycles) from the current period
  period = abs(midday.index(period)-(cycles % 2)) # period is an index in midday
  passed = (period + cycles) // 2  # 'passed' describes the number of days passed 
  
  if hour == 0: # basically an edge case created from my modulus calculations
    hour = 12

  if minutes < 10: # standardizing time format, 12:1 -> 12:01
    minutes = f'0{minutes}'

  # Day of Week:
  # In one line, we can determine what the current day of the week is by index, add the number of days passed, and get the new day of the week
  # The modulus operation will scale the new day, so day 10 -> day 3, day 145 -> day 5, etc.
  if day:
      week = ('Sunday','Monday','Tuesday','Wednesday','Thursday','Friday','Saturday')
      new_day = f', {week[(week.index(day.capitalize()) + passed) % 7]}' 
  
  # Just some string formatting 
  if passed == 1:
      later = ' (next day)'
  elif passed != 0:
      later = f' ({passed} days later)'
      
  return f'{hour}:{minutes} {midday[period]}{new_day}{later}'

Here is a link to the code as well: https://replit.com/@ROOPESH-J/boilerplate-time-calculator#time_calculator.py

Thanks in advance for any suggestions and feel free to ask questions if something is unclear.

Edit: I apologize for the bad formatting in the code block.
Edit: Cleaning up the comments.

2

faster than what?

I added spoiler tags to your solution.

3

Please do not revert moderator edits. Thank you.

4

Oh shoot, I forgot about the spoiler tag. Sorry about that. For my reasoning on using a tuple, I mean faster than creating and using a list. I know its completely insignificant for this program size but I thought I’d would just build that awareness for myself

5

I meant to revert my own edit, my bad

6

‘Faster’ should be a concern when you know something is a performance critical piece of code. Clearer should be the default criteria for picking an algorithm or data structure unless you know the choice will have an important performance impact.

In this case, there is nothing faster about using a tuple or a list or any other basic data structure for a pair of short strings, so I wouldn’t make a misleading comment about ‘faster’.

7

Yeah in hindsight, that makes a lot of sense. I can see how the comment could be misleading, will change that. Thank you!

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