Ordering a list of tuples in python

Question

I have a list, my_list:

[['28/02/2014, apples']
['09/07/2014, oranges']
['22/04/2014, bananas']
['14/03/2014, tomatoes']]

which I am trying to order by date. This is the code I am using:

def display(my_list):
    for item in my_list:
        x = ([item[0] + ": " + item[1]])
        print x

I've tried applying different forms of sorting to x (sorted, x.sort(), etc) but nothing seems to work. How can I get the list to sort by date, from earliest to latest?

I'm very confused by what your code actually looks like. That first block of code is not "a list of tuples", it is four separate single-element lists each containing exactly one string. — Two-Bit Alchemist, Apr 08 '14 at 17:13
Sorry for the confusion - the list is actually what the function displays at the moment, without sorting. I'll try and edit it to make it a bit more clear. — Chrystael, Apr 08 '14 at 17:15
possible duplicate of [How to sort (list/tuple) of lists/tuples?](http://stackoverflow.com/questions/3121979/how-to-sort-list-tuple-of-lists-tuples) — Gilles 'SO- stop being evil', Apr 08 '14 at 17:21
@Gilles I doubt it is a duplicate - there are dates inside of strings - it makes the question different. — alecxe, Apr 08 '14 at 17:22

alecxe · Accepted Answer · 2014-04-08T17:20:55.017

You can use sorted() with applying a key function that takes the first item from every sublist, splits it by : and converts the part before the colon to datetime using strptime():

>>> from datetime import datetime
>>> l = [['28/02/2014: apples'], ['09/07/2014: oranges'], ['22/04/2014: bananas'], ['14/03/2014: tomatoes']]
>>> sorted(l, key=lambda x: datetime.strptime(x[0].split(':')[0], '%d/%m/%Y'))
[['28/02/2014: apples'],  
 ['14/03/2014: tomatoes'], 
 ['22/04/2014: bananas'], 
 ['09/07/2014: oranges']]

Ordering a list of tuples in python

1 Answers1