Sort Dictionary Keys in natural order [duplicate]

Question

I want to sort dictionary keys in a "natural order". If I have a dictionary with the keys

    d = {"key1" : object, "key11" : object, "key2" : object, "key22" : object", "jay1" : object, "jay2" : object}

I want to sort this dictionary so the result is:

    d = { "jay1" : object, "jay2" : object, "key_1" : object, "key_2" : object, "key_11" : object, "key_22" : object"}

Answer 1

You can change your dict into OrderedDict:

import collections, re

d = {"key1" : 'object', "key11" : 'object', "key2" : 'object', "key22" : 'object', "jay1" : 'object', "jay2" : 'object'}


my_fun = lambda k,v: [k, int(v)]

d2 = collections.OrderedDict(sorted(d.items(), key=lambda t: my_fun(*re.match(r'([a-zA-Z]+)(\d+)',t[0]).groups())))

print(d2)
#reslt: OrderedDict([('jay1', 'object'), ('jay2', 'object'), ('key1', 'object'), ('key11', 'object'), ('key2', 'object'), ('key22', 'object')])

Basically, what is happening here, that I split the strings into 'string' part and number part. Number part is changed to int, and the sorting happens using these two values.

Answer 2

As others have said, dictionaries are not ordered. However, if you want to iterate through these keys in a natural order, you could do something like the following:

d = {"key1" : object, "key11" : object, "key2" : object, "key22" : object, "jay1" : object, "jay2" : object}
sortedKeys = sorted(d.keys())
print sortedKeys
for key in sortedKeys:
    print d[key]