Ruby to python one-liner conversion

Question

I have a little one-liner in my Rails app that returns a range of copyright dates with an optional parameter, e.g.:

def copyright_dates(start_year = Date.today().year)
    [start_year, Date.today().year].sort.uniq.join(" - ")
end

I'm moving the app over to Django, and while I love it, I miss a bit of the conciseness. The same method in Python looks like:

def copyright_dates(start_year = datetime.datetime.today().year):
    years = list(set([start_year, datetime.datetime.today().year]))
    years.sort()
    return " - ".join(map(str, years))

It's been years since I've touched Python, so I'm betting there's an easier way to do it. Any ideas?

EDIT: I know lists and sets are a bit of overkill, but I want the following output assuming the code is run in 2009:

copyright_dates()     # '2009'
copyright_dates(2007) # '2007 - 2009'
copyright_dates(2012) # '2009 - 2012'

Answer 1

from datetime import datetime

def copyright_dates(start_year = datetime.now().year):
    return " - ".join(str(y) for y in sorted(set([start_year, datetime.now().year])))

Answer 2

Lists and sets seem to be overkill to me.

How about this:

def copyright_dates(start=datetime.datetime.today().year):
    now = datetime.datetime.today().year
    return (start==now and str(now) or "%d - %d" % (min(start, now), max(start, now)))

Answer 3

Watch out for the default parameter which is evaluated once. So if your web application runs over 12/31/09 without a restart, you won't get the expected output.

Try:

def copy(start=None):
    start, curr = start if start else datetime.today().year, datetime.today().year
    return str(start) if start == curr else '%d - %d' % tuple(sorted([start, curr]))