#pragma section-numbers off = Sorting Lists of Dictionaries = Frequently you want to sort a list of dictionaries, based on some particular key. For example: {{{ #!python a = {"key1": 5 , "key2": 8, "key3": 2} b = {"key1": 7 , "key2": 4, "key3": 9} c = {"key1": 6 , "key2": 1, "key3": 1} 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. {{{ #!python sort_on = "key2" decorated = [(dict_[sort_on], dict_) for dict_ in undecorated] decorated.sort() 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')) }}} This will sort on arbitrary multiple columns of the dictionary. {{{ #!python def multikeysort(items, columns): from operator import itemgetter comparers = [ ((itemgetter(col[1:].strip()), -1) if col.startswith('-') else (itemgetter(col.strip()), 1)) for col in columns] def comparer(left, right): for fn, mult in comparers: result = cmp(fn(left), fn(right)) if result: return mult * result else: return 0 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']) }}} == See Also == * [[HowTo/Sorting]]