sorting dictionary python 3

Question

I'm working on python 3.2.2. Breaking my head more than 3 hours to sort a dictionary by it's keys. I managed to make it a sorted list with 2 argument members, but can not make it a sorted dictionary in the end.

This is what I've figured:

myDic={10: 'b', 3:'a', 5:'c'} sorted_list=sorted(myDic.items(), key=lambda x: x[0]) 

But no matter what I can not make a dictionary out of this sorted list. How do I do that? Thanks!

Answer 1

A modern and fast solution, for Python 3.7. May also work in some interpreters of Python 3.6.

TLDR

To sort a dictionary by keys use:

sorted_dict = {k: disordered[k] for k in sorted(disordered)} 

Almost three times faster than the accepted answer; probably more when you include imports.

Comment on the accepted answer

The example in the accepted answer instead of iterating over the keys only - with key parameter of sorted() or the default behaviour of dict iteration - iterates over tuples (key, value), which suprisingly turns out to be much slower than comparing the keys only and accessing dictionary elements in a list comprehension.

How to sort by key in Python 3.7

The big change in Python 3.7 is that the dictionaries are now ordered by default.

• You can generate sorted dict using dict comprehensions.
• Using OrderedDict might still be preferable for the compatibility sake.
• Do not use sorted(d.items()) without key.

See:

disordered = {10: 'b', 3: 'a', 5: 'c'}  # sort keys, then get values from original - fast sorted_dict = {k: disordered[k] for k in sorted(disordered)}  # key = itemgetter - slower from operator import itemgetter key = itemgetter(0) sorted_dict = {k: v for k, v in sorted(disordered.items(), key=key)}  # key = lambda - the slowest key = lambda item: item[0] sorted_dict = {k: v for k in sorted(disordered.items(), key=key)}  

Timing results:

Best for {k: d[k] for k in sorted(d)}: 7.507327548999456 Best for {k: v for k, v in sorted(d.items(), key=key_getter)}: 12.031082626002899 Best for {k: v for k, v in sorted(d.items(), key=key_lambda)}: 14.22885995300021  Best for dict(sorted(d.items(), key=key_getter)): 11.209122000000207 Best for dict(sorted(d.items(), key=key_lambda)): 13.289728325995384 Best for dict(sorted(d.items())): 14.231471302999125  Best for OrderedDict(sorted(d.items(), key=key_getter)): 16.609151654003654 Best for OrderedDict(sorted(d.items(), key=key_lambda)): 18.52622927199991 Best for OrderedDict(sorted(d.items())): 19.436101284998585 

Testing code:

from timeit import repeat  setup_code = """ from operator import itemgetter from collections import OrderedDict import random random.seed(0) d = {i: chr(i) for i in [random.randint(0, 120) for repeat in range(120)]} key_getter = itemgetter(0) key_lambda = lambda item: item[0] """  cases = [     # fast     '{k: d[k] for k in sorted(d)}',     '{k: v for k, v in sorted(d.items(), key=key_getter)}',     '{k: v for k, v in sorted(d.items(), key=key_lambda)}',     # slower     'dict(sorted(d.items(), key=key_getter))',     'dict(sorted(d.items(), key=key_lambda))',     'dict(sorted(d.items()))',     # the slowest      'OrderedDict(sorted(d.items(), key=key_getter))',     'OrderedDict(sorted(d.items(), key=key_lambda))',     'OrderedDict(sorted(d.items()))', ]  for code in cases:     times = repeat(code, setup=setup_code, repeat=3)     print(f"Best for {code}: {min(times)}")