How Can I Sort a Dictionary by Key?

How can I easily go from {2:3, 1:89, 4:5, 3:0} to {1:89, 2:3, 3:0, 4:5}? All the posts I could find about sorting dictionaries use the built-in sorted function, e.g.

>>> x = {2:3, 1:89, 4:5, 3:0}
>>> sorted(x.items())
[(1, 89), (2, 3), (3, 0), (4, 5)]

This is not what I want - I want the result to be a dictionary.

9

32 Answers

Note: for Python 3.7+, see this answer

Standard Python dictionaries are unordered (until Python 3.7). Even if you sorted the (key,value) pairs, you wouldn't be able to store them in a dict in a way that would preserve the ordering.

The easiest way is to use OrderedDict, which remembers the order in which the elements have been inserted:

In [1]: import collections

In [2]: d = {2:3, 1:89, 4:5, 3:0}

In [3]: od = collections.OrderedDict(sorted(d.items()))

In [4]: od
Out[4]: OrderedDict([(1, 89), (2, 3), (3, 0), (4, 5)])

Never mind the way od is printed out; it'll work as expected:

In [11]: od[1]
Out[11]: 89

In [12]: od[3]
Out[12]: 0

In [13]: for k, v in od.iteritems(): print k, v
   ....: 
1 89
2 3
3 0
4 5

Python 3

For Python 3 users, one needs to use the .items() instead of .iteritems():

In [13]: for k, v in od.items(): print(k, v)
   ....: 
1 89
2 3
3 0
4 5
15

Dictionaries themselves do not have ordered items as such, should you want to print them etc to some order, here are some examples:

In Python 2.4 and above:

mydict = {'carl':40,
          'alan':2,
          'bob':1,
          'danny':3}

for key in sorted(mydict):
    print "%s: %s" % (key, mydict[key])

gives:

alan: 2
bob: 1
carl: 40
danny: 3

(Python below 2.4:)

keylist = mydict.keys()
keylist.sort()
for key in keylist:
    print "%s: %s" % (key, mydict[key])

Source:

5

For CPython/PyPy 3.6, and any Python 3.7 or higher, this is easily done with:

>>> d = {2:3, 1:89, 4:5, 3:0}
>>> dict(sorted(d.items()))
{1: 89, 2: 3, 3: 0, 4: 5}
6

From Python's collections library documentation:

>>> from collections import OrderedDict

>>> # regular unsorted dictionary
>>> d = {'banana': 3, 'apple':4, 'pear': 1, 'orange': 2}

>>> # dictionary sorted by key -- OrderedDict(sorted(d.items()) also works
>>> OrderedDict(sorted(d.items(), key=lambda t: t[0]))
OrderedDict([('apple', 4), ('banana', 3), ('orange', 2), ('pear', 1)])

>>> # dictionary sorted by value
>>> OrderedDict(sorted(d.items(), key=lambda t: t[1]))
OrderedDict([('pear', 1), ('orange', 2), ('banana', 3), ('apple', 4)])

>>> # dictionary sorted by length of the key string
>>> OrderedDict(sorted(d.items(), key=lambda t: len(t[0])))
OrderedDict([('pear', 1), ('apple', 4), ('orange', 2), ('banana', 3)])
2

There are a number of Python modules that provide dictionary implementations which automatically maintain the keys in sorted order. Consider the sortedcontainers module which is pure-Python and fast-as-C implementations. There is also a performance comparison with other popular options benchmarked against one another.

Using an ordered dict is an inadequate solution if you need to constantly add and remove key/value pairs while also iterating.

>>> from sortedcontainers import SortedDict
>>> d = {2:3, 1:89, 4:5, 3:0}
>>> s = SortedDict(d)
>>> s.items()
[(1, 89), (2, 3), (3, 0), (4, 5)]

The SortedDict type also supports indexed location lookups and deletion which isn't possible with the built-in dict type.

>>> s.iloc[-1]
4
>>> del s.iloc[2]
>>> s.keys()
SortedSet([1, 2, 4])
0

Simply:

d = {2:3, 1:89, 4:5, 3:0}
sd = sorted(d.items())

for k,v in sd:
    print k, v

Output:

1 89
2 3
3 0
4 5
3

Python dictionary was unordered before Python 3.6. In CPython implementation of Python 3.6, dictionary keeps the insertion order. From Python 3.7, this will become a language feature.

In changelog of Python 3.6 ():

The order-preserving aspect of this new implementation is considered an implementation detail and should not be relied upon (this may change in the future, but it is desired to have this new dict implementation in the language for a few releases before changing the language spec to mandate order-preserving semantics for all current and future Python implementations; this also helps preserve backwards-compatibility with older versions of the language where random iteration order is still in effect, e.g. Python 3.5).

In document of Python 3.7 ():

Performing list(d) on a dictionary returns a list of all the keys used in the dictionary, in insertion order (if you want it sorted, just use sorted(d) instead).

So unlike previous versions, you can sort a dict after Python 3.6/3.7. If you want to sort a nested dict including the sub-dict inside, you can do:

test_dict = {'a': 1, 'c': 3, 'b': {'b2': 2, 'b1': 1}}

def dict_reorder(item):
    return {k: dict_reoder(v) if isinstance(v, dict) else v for k, v in sorted(item.items())}

reordered_dict = dict_reorder(test_dict)
1

As others have mentioned, dictionaries are inherently unordered. However, if the issue is merely displaying dictionaries in an ordered fashion, you can override the __str__ method in a dictionary subclass, and use this dictionary class rather than the builtin dict. Eg.

class SortedDisplayDict(dict):
   def __str__(self):
       return "{" + ", ".join("%r: %r" % (key, self[key]) for key in sorted(self)) + "}"


>>> d = SortedDisplayDict({2:3, 1:89, 4:5, 3:0})
>>> d
{1: 89, 2: 3, 3: 0, 4: 5}

Note, this changes nothing about how the keys are stored, the order they will come back when you iterate over them etc, just how they're displayed with print or at the python console.

Found another way:

import json
print json.dumps(d, sort_keys = True)

upd:
1. this also sorts nested objects (thanks @DanielF).
2. python dictionaries are unordered therefore this is sutable for print or assign to str only.

2

In Python 3.

>>> D1 = {2:3, 1:89, 4:5, 3:0}
>>> for key in sorted(D1):
    print (key, D1[key])

gives

1 89
2 3
3 0
4 5
0

An easy way to do this:

d = {2:3, 1:89, 4:5, 3:0}

s = {k : d[k] for k in sorted(d)}

s
Out[1]: {1: 89, 2: 3, 3: 0, 4: 5} 
2

There are plenty of answers here already showcasing popular ways to sort a Python dictionary. I thought I'd add a few more less-obvious ways for those coming here from Google looking for non-standard ideas.

Sample Dictionary: d = {2: 'c', 1: 'b', 0: 'a', 3: 'd'}

Elena Rostova

Elena Rostova

Lead Health, Wellness & Medical Journalist

Elena Rostova holds a Master's degree in Public Health Journalism. She covers groundbreaking medical research, holistic wellness trends, mental health awareness, and nutritional science.