Sorting Lists of Dictionaries
Frequently you want to sort a list of dictionaries, based on some particular key.
For example:
1 a = {"key1": 5 , "key2": 8, "key3": 2}
2 b = {"key1": 7 , "key2": 4, "key3": 9}
3 c = {"key1": 6 , "key2": 1, "key3": 1}
4 undecorated = [a, b, c] # how do you sort this list?
There are many ways to do this. Here's the fastest way to do it, as it avoids using a custom comparison function, instead using builtin comparisons. This is the decorate-sort-undecorate pattern, or the Schwartzian transform if you're coming from Perl.
1 sort_on = "key2"
2 decorated = [(dict_[sort_on], dict_) for dict_ in undecorated]
3 decorated.sort()
4 result = [dict_ for (key, dict_) in decorated]
(The variable was named dict_ because dict is already a builtin.)
Starting with Py2.4 the list.sort() method provides a key= argument for doing the transform in a single step. The new sorted() built-in function goes a step further and encapsulates making a new sorted list while leaving the original intact. Also, the new operator.itemgetter() function helps by constructing functions for key access:
>>> from operator import itemgetter
>>> result = sorted(undecorated, key=itemgetter('key2'))