Python Cookbook

Sorting a Dictionary

Credit: Alex Martelli, Raymond Hettinger

Problem

You want to sort a dictionary. Because sorting is a concept that only makes sense for sequences, this presumably means that you want a sequence of the values of the dictionary in the order obtained by sorting the keys.

Solution

The simplest approach is to sort the items (i.e., key/value pairs), then pick just the values:

def sortedDictValues1(adict):
    items = adict.items(  )
    items.sort(  )
    return [value for key, value in items]

However, an alternative implementation that sorts just the keys, then uses them to index into the dictionary to build the result, happens to run more than twice as fast (for a dictionary of a few thousand entries) on my system:

def sortedDictValues2(adict):
    keys = adict.keys(  )
    keys.sort(  )
    return [adict[key] for key in keys]

A further small speed-up (15% on my system) is to perform the last step by mapping a bound method. map is often marginally faster than a list comprehension when no lambda is involved:

def sortedDictValues3(adict):
    keys = adict.keys(  )
    keys.sort(  )
    return map(adict.get, keys)

A really tiny extra speed-up (about 3% on my system) is available in Python 2.2 by using adict._ _getitem_ _ rather than adict.get in this latest, bound-method version.

Discussion

The concept of sorting applies only to a collection that has order—in other words, a sequence. A mapping, such as a dictionary, has no order, so it cannot be sorted. And yet, “How do I sort a dictionary?” is a frequent question ...