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How can I sort a dictionary by key?


Question

What would be a nice way to go from {2:3, 1:89, 4:5, 3:0} to {1:89, 2:3, 3:0, 4:5}?
I checked some posts but they all use the "sorted" operator that returns tuples.

2015/05/01
1
981
5/1/2015 10:57:52 PM

Accepted Answer

Standard Python dictionaries are unordered. 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
2016/03/12
972
3/12/2016 11:38:38 AM

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: http://www.saltycrane.com/blog/2007/09/how-to-sort-python-dictionary-by-keys/

2014/04/15

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)])
2014/06/09

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}
2019/08/27

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])
2014/03/28

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
2015/11/12

Source: https://stackoverflow.com/questions/9001509
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