Question

I have a dictionary with the following structure:

{'ONE' : (4, 6, 9), 'TWO' : (3, 8, 10)}

I’d like to sort this dictionary by the 3rd value of each tuple. I can’t really think of a way to accomplish this with the data setup as a dictionary, but it should be fairly easy if I can convert the data to a nested list like the following:

[['ONE', 4, 6, 9], ['TWO', 3, 8, 10]]

I’m looking for the most efficient code for accomplishing this. If there is a way to sort this dictionary without converting it, that would be ideal. If not, any assistance with the conversion from dictionary to nested list would be greatly appreciated. Thanks.

Answer 1

Use the key argument of the built-in sorted function:

>>> d = {'ONE' : (4, 6, 9), 'TWO' : (3, 8, 10), 'FOUR': (2, 5, 8)}
>>> sorted(d.iteritems(), key=lambda i: i[1][2])
[('FOUR', (2, 5, 8)), ('ONE', (4, 6, 9)), ('TWO', (3, 8, 10))]

EDIT

If some of the values are ints rather than tuples, then something like this should work:

>>> d = {'ONE' : (4, 6, 9), 'TWO' : (3, 8, 10), 'FOUR': (2, 5, 8), 'THREE': 0}
>>> sorted(d.iteritems(),
...        key=lambda i: i[1][2] if isinstance(i[1], tuple) else i[1])
[('THREE', 0), ('FOUR', (2, 5, 8)), ('ONE', (4, 6, 9)), ('TWO', (3, 8, 10))]

However, in the long run, it’s probably better to normalize the data so that the values all have the same format.