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
– Two-Bit Alchemist, Commented Apr 8, 2014 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
– Chrystael, Commented Apr 8, 2014 at 17:15
possible duplicate of How to sort (list/tuple) of lists/tuples? — Gilles 'SO- stop being evil'
– Gilles 'SO- stop being evil', Commented Apr 8, 2014 at 17:21
@Gilles I doubt it is a duplicate - there are dates inside of strings - it makes the question different. — alecxe
– alecxe, Commented Apr 8, 2014 at 17:22

alecxe · Accepted Answer · 2014-04-08 17:20:55Z

6

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']]

edited Apr 8, 2014 at 17:20

answered Apr 8, 2014 at 17:15

alecxe

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