Sorting Lists of Dictionaries
Frequently you want to sort a list of dictionaries, based on some particular key.
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.
(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'))
This will sort on arbitrary multiple columns of the dictionary.
1 def multikeysort(items, columns): 2 from operator import itemgetter 3 comparers = [ ((itemgetter(col[1:].strip()), -1) if col.startswith('-') else (itemgetter(col.strip()), 1)) for col in columns] 4 def sign(a, b): 5 if a < b: return -1 6 elif a > b: return 1 7 else: return 0 8 def comparer(left, right): 9 for fn, mult in comparers: 10 result = sign(fn(left), fn(right)) 11 if result: 12 return mult * result 13 else: 14 return 0 15 return sorted(items, cmp=comparer)
You can call it like this:
>>> result = multikeysort(undecorated, ['key1', 'key2', 'key3'])
Column names preceded by '-' are sorted in descending order:
>>> result = multikeysort(undecorated, ['-key1', '-key2', '-key3'])