Python: Converting ('Monday', 'Tuesday', 'Wednesday') to 'Monday to Wednesday'

Question

I'm getting a sequence of day of the week. Python code of what I want to do:

def week_days_to_string(week_days):
    """
    >>> week_days_to_string(('Sunday', 'Monday', 'Tuesday'))
    'Sunday to Tuesday'
    >>> week_days_to_string(('Monday', 'Wednesday'))
    'Monday and Wednesday'
    >>> week_days_to_string(('Sunday', 'Wednesday', 'Thursday'))
    'Sunday, Wednesday, Thursday'
    """
    if len(week_days) == 2:
       return '%s and %s' % weekdays
    elif week_days_consecutive(week_days):
       return '%s to %s' % (week_days[0], week_days[-1])
    return ', '.join(week_days)

I just need the week_days_consecutive function (the hard part heh).

Any ideas how I could make this happen?

Clarification:

My wording and examples caused some confusion. I do not only want to limit this function to the work week. I want to consider all days of the week (S, M, T, W, T, F). My apologies for not being clear about that last night. Edited the body of the question to make it clearer.

Edit: Throwing some wrenches into it

Wraparound sequence:

>>> week_days_to_string(('Sunday', 'Monday', 'Tuesday', 'Saturday'))
'Saturday to Tuesday'

And, per @user470379 and optional:

>>> week_days_to_string(('Monday, 'Wednesday', 'Thursday', 'Friday'))
'Monday, Wednesday to Friday'

Answer 1

I would approach this problem by:

• Creating a dict mapping day names to their sequential index
• Converting my input day names to their sequential indices
• Looking at the resulting input indices and asking if they are sequential

Here's how you can do that, using calendar.day_name, range and some for comprehensions:

day_indexes = {name:i for i, name in enumerate(calendar.day_name)}
def weekdays_consecutive(days):
    indexes = [day_indexes[d] for d in days]
    expected = range(indexes[0], indexes[-1] + 1)
    return indexes == expected

A few other options, depending on what you need:

• If you need Python < 2.7, instead of the dict comprehension, you can use:

  day_indexes = dict((name, i) for i, name in enumerate(calendar.day_name))
  

• If you don't want to allow Saturday and Sunday, just trim off the last two days:

  day_indexes = ... calendar.day_name[:-2] ...
  

• If you need to wrap around after Sunday, it's probably easiest to just check that each item is one more than the previous item, but working in modulo 7:

  def weekdays_consecutive(days):
      indexes = [day_indexes[d] for d in days]
      return all(indexes[i + 1] % 7 == (indexes[i] + 1) % 7
                 for i in range(len(indexes) - 1))
  

Update: For the extended problem, I would still stick with they day-to-index dict, but instead I would:

• Find all the indexes where a run of sequential days stops
• Wrap the days around if necessary to get the longest possible sequence of days
• Group the days into their sequential spans

Here's code to do this:

def weekdays_to_string(days):
    # convert days to indexes
    day_indexes = {name:i for i, name in enumerate(calendar.day_name)}
    indexes = [day_indexes[d] for d in days]

    # find the places where sequential days end
    ends = [i + 1
            for i in range(len(indexes))
            if (indexes[(i + 1) % len(indexes)]) % 7 !=
               (indexes[(i) % len(indexes)] + 1) % 7]

    # wrap the days if necessary to get longest possible sequences
    split = ends[-1]
    if split != len(days):
        days = days[split:] + days[:split]
        ends = [len(days) - split + end for end in ends]

    # group the days in sequential spans
    spans = [days[begin:end] for begin, end in zip([0] + ends, ends)]

    # format as requested, with "to", "and", commas, etc.
    words = []
    for span in spans:
        if len(span) < 3:
            words.extend(span)
        else:
            words.append("%s to %s" % (span[0], span[-1]))
    if len(days) == 1:
        return words[0]
    elif len(days) == 2:
        return "%s and %s" % tuple(words)
    else:
        return ", ".join(words)

You might also try the following instead of that last if/elif/else block to get an "and" between the last two items and commas between everything else:

    if len(words) == 1:
        return words[0]
    else:
        return "%s and %s" % (", ".join(words[:-1]), words[-1])

That's a little different from the spec, but prettier in my eyes.