Sorting a dictionary with lists as values, according to an element from the list

Solution 1:

Here is one way to do this:

>>> sorted(myDict.items(), key=lambda e: e[1][2])
[('item2', [8, 2, 3]), ('item1', [7, 1, 9]), ('item3', [9, 3, 11])]

The key argument of the sorted function lets you derive a sorting key for each element of the list.

To iterate over the keys/values in this list, you can use something like:

>>> for key, value in sorted(myDict.items(), key=lambda e: e[1][2]):
...   print key, value
... 
item2 [8, 2, 3]
item1 [7, 1, 9]
item3 [9, 3, 11]

Solution 2:

You stated two quite different wants:

1. "What I want to do is sort a dictionary of lists ..."
2. "I want to be able to iterate through the dictionary in order of ..."

The first of those is by definition impossible -- to sort something implies a rearrangement in some order. Python dictionaries are inherently unordered. The second would be vaguely possible but extremely unlikely to be implemented.

What you can do is

1. Take a copy of the dictionary contents (which will be quite unordered)
2. Sort that
3. Iterate over the sorted results -- and you already have two solutions for that. By the way, the solution that uses "key" instead of "cmp" is better; see sorted

"the third item in the list" smells like "the third item in a tuple" to me, and "e[1][2]" just smells :-) ... you may like to investigate using named tuples instead of lists; see named tuple factory

If you are going to be doing extract/sort/process often on large data sets, you might like to consider something like this, using the Python-supplied sqlite3 module:

create table ex_dict (k text primary key, v0 int, v1 int, v2 int);
insert into ex_dict values('item1', 7, 1, 9);
-- etc etc 
select * from ex_dict order by v2;