# 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'))```

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 comparer(left, right):
5         for fn, mult in comparers:
6             result = cmp(fn(left), fn(right))
7             if result:
8                 return mult * result
9         else:
10             return 0
11     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'])`