Sorting a nested list in python

You have a list of lists, so you need to sort each sublist if you want to order the sublists.

a_sorted = [sorted(sublist) for sublist in a]

Obviously, you can run sorted() on the list of sorted lists if you want the outer list to be sorted too.

You can pass the sorted() builtin a key argument - a function which takes the list item and returns the value to sort on. It's a little unclear exactly how you want to sort the list, you could clarify, although with this you should be able to work out a solution.

Your data structure seems a little odd, however. Where you have pairs in the form {key: value} you would normally be better off with a tuple: (key, value), or a single, larger dict that contains all of your pairs as keys and values, e.g:

a = [{'aa': 2L, 'mm': 7l}, {'aa': 2L, 'mm': 5L}, {'aa': 2L, 'mm': 9L}, {'aa': 2L, 'mm': 3L}]

In this case, we can use sorted(a, key=itemgetter("mm")) - using operator.itemgetter() - to sort on the value of 'mm'.

Or, if you need your empty pairs (without using keys to None, for example), as tuples:

a = [[('aa', 2L), (,), ('mm', 7L), (,), (,)], [('aa', 2L), (,), ('mm', 5L), (,), (,)], [('aa', 2L), (,), ('mm', 9L), (,), (,)], [('aa', 2L), (,), ('mm', 3L), (,), (,)]]

Here we can do a similar thing sorted(a, key=lambda sublist: sublist[2][1]) - we use lambda to make a quick function to extract the second item in the third item in the sublist.

If you wanted to keep your data structure as is - if, for example, you plan to expand the dictionaries with more content, then a similar plan would work sorted(a, key=lambda sublist: sublist[2]["mm"]) - this time using 'mm' to access in the dict.