How to sort a dictionary having keys as a string of numbers in Python

Question

I have a dictionary:

a = {'100':12,'6':5,'88':3,'test':34, '67':7,'1':64 }

I want to sort this dictionary with respect to key so it looks like:

a = {'1':64,'6':5,'67':7,'88':3, '100':12,'test':34 }

Answer 1

Like everyone else has pointed out, dictionaries have their own ordering and you can't just sort them like you would a list.

One thing I would like to add is that, if you just want to go through the elements of a dictionary in sorted order, that's just:

for k in sorted(a):
    print k, a[k] # or whatever.

If you'd rather have a list comprehension (per Alex):

sortedlist = [(k, a[k]) for k in sorted(a)]

I would like to point out that Alex's use of key=int won't work with your example because one of your keys is 'test' . If you really want he numbers sorted before the non-numerics, you'll have to pass in a cmp function:

def _compare_keys(x, y):
    try:
        x = int(x)
    except ValueError:
        xint = False
    else:
        xint = True
    try:
        y = int(y)
    except ValueError:
        if xint:
            return -1
        return cmp(x.lower(), y.lower())
        # or cmp(x, y) if you want case sensitivity.
    else:
        if xint:
            return cmp(x, y)
        return 1

for k in sorted(a, cmp=_compare_keys):
    print k, a[k] # or whatever.

Or maybe you know enough about your keys to write a function to convert them into a string (or other object) that sorts right:

# Won't work for integers with more than this many digits, or negative integers.
MAX_DIGITS = 10
def _keyify(x):
    try:
        xi = int(x)
    except ValueError:
        return 'S{0}'.format(x)
    else:
        return 'I{0:0{1}}'.format(xi, MAX_DIGITS)

for k in sorted(a, key=_keyify):
    print k, a[k] # or whatever.

This would be much faster than using a cmp function.

Answer 2

You cannot sort a dict in Python as the dict type is inherently unordered. What you can do is sort the items before you used them using the sorted() built in function. You will also need a helper function to distinguish between your numerical and string keys:

def get_key(key):
    try:
        return int(key)
    except ValueError:
        return key
a = {'100':12,'6':5,'88':3,'test':34, '67':7,'1':64 }
print sorted(a.items(), key=lambda t: get_key(t[0]))

However in Python 3.1 (and 2.7) the collections module contains the collections.OrderedDict type that can be used to achieve the effect you want like below:

def get_key(key):
    try:
        return int(key)
    except ValueError:
        return key
a = {'100':12,'6':5,'88':3,'test':34, '67':7,'1':64 }
b = collections.OrderedDict(sorted(a.items(), key=lambda t: get_key(t[0])))
print(b)

Answer 3

9 years ago I posted arecipe that starts

Dictionaries can't be sorted -- a mapping has no ordering!

and shows how to get sorted lists out of a dict's keys and values.

With today's Python, and your expressed-plus-implied specs, I'd suggest:

import sys

def asint(s):
    try: return int(s), ''
    except ValueError: return sys.maxint, s

sortedlist = [(k, a[k]) for k in sorted(a, key=asint)]

the key=asint is what tells sorted to treat those string keys as integers for sorting purposes, so that eg '2' sorts between '1' and '12' , rather than after them both -- that's what you appear to require, as well as having all non-all-digits keys sort after all all-digits ones. If you need to also deal with all-digits key strings that express integers larger than sys.maxint, it's a bit trickier, but still doable:

class Infinity(object):
    def __cmp__(self, other): return 0 if self is other else 1
infinite = Infinity()
def asint(s):
    try: return int(s), ''
    except ValueError: return infinite, s

In general, you can get better answers faster if you specify your exact requirements more precisely from the start;-).

Answer 4

Dictionaries are unordered. You cannot sort one like you show because the resultant a is a dict, and dicts do not have order.

If you want, say, a list a list of the keys in sorted order, you can use code like

>>> def my_key(dict_key):
...     try:
...         return int(dict_key)
...     except ValueError:
...         return dict_key
...
>>> sorted(a, key=my_key)
['1', '6', '67', '88', '100', 'test']

This relies on the stupid Python behavior that instances of str are always greater than instances of int . (The behaviour is fixed in Python 3.) In an optimal design, the keys of your dict would be things you could compare sanely and you wouldn't mix in strings representing numbers with strings representing words.

If you want to keep the keys in always-sorted order, you can use the bisect module or implement a mapping that relies on a tree data structure. The bisect module does not accept a key argument like the sorting stuff because this would be potentially inefficient; you would use the decorate–use–undecorate pattern if you chose to use bisect , keeping a sorted list that depends on the result of the key function.

Answer 5

If you install my blist package, it includes a sorteddict type. Then you could simply:

from blist import sorteddict

def my_key(dict_key):
       try:
              return int(dict_key)
       except ValueError:
              return dict_key

a = {'100':12,'6':5,'88':3,'test':34, '67':7,'1':64 }
print sorteddict(my_key, **a).keys()

Output:

['1', '6', '67', '88', '100', 'test']

Answer 6

what you could do is also the following:

sorted_keys = sorted(list(a.keys()), key = lambda x: (len(x),x))
sorted_dict = {k:a[k] for k in sorted_keys}

The most important part is key = lambda x: (len(x),x) which I took from here