python - selecting the earliest entry in a list -

if have list people's names , dates, , want keep entry earliest date per person how do that? want final list alphabetical last name, first name , contain entry earliest date @ end.

here example of list , tried, gave me same list again.

l1=['smith, john, 1994', 'smith, john, 1996', 'smith, john, 1998', 'smith, joan, 1993', 'smith, joan, 1995', 'smith, jack, 1989', 'smith, jack, 1991', 'jones, adam, 2000', 'jones, adam, 1998', 'jones, sarah, 2002', 'jones, sarah, 2005', 'brady, tom, 2001', 'brady, tonya, 2002']  l1.sort()  l2= []  item in l1:     if item.split(',')[:2] not in l2:         l2.append(item) 

the final product should like:

l2=['brady, tom, 2001', 'brady, tonya, 2002', 'jones, adam, 1998', 'jones, sarah, 2002', 'smith, jack, 1989', 'smith, joan, 1993', 'smith, john, 1994'] 

any or insight appreciated!

try

l1.sort() [next(j) i, j in itertools.groupby(l1, lambda x: x.rsplit(",", 1)[0])] 

your code not work since searching l2 item.split(',')[:2], name. strings in list consist of name , year -- that's why not in yields true.