Explicitly sorting a dictionary by a sub-dictionary subkey into an ordered list.

sorting_example_by_subdict_key

# Dictionaries are inherently orderless, but lists are ordered!

import collections

d = { '123': { 'key1': 3, 'key2': 11, 'key3': 3 }, 
	  '124': { 'key1': 6, 'key2': 56, 'key3': 6 }, 
	  '125': { 'key1': 7, 'key2': 44, 'key3': 9 } }

results_list = []
sort_by_dict = {} 
# this is going to hold the original index 
# and the order like this {original_dict_index: order}

# Make a sorting dictionary full of the original key index and 
# value of subkey3
value_dict = {}
for key in d:
	value = d[key]
	for subkey in value:
		subvalue = value[subkey]
		if subkey == 'key3':
			original_index = key
			subkey3value = subvalue
			sort_by_dict[original_index] = subkey3value

# This is handles duplicate values with a number of occurences count
number_of_occurences = collections.defaultdict(int)
for key in sort_by_dict:
	value = sort_by_dict[key]
	number_of_occurences[value] += 1

# This is a sorted list not a dict, hence the i below ;)
sorted_number_occurences = sorted(number_of_occurences, key=number_of_occurences.get, reverse=True)

i = 0
for key in sorted_number_occurences:
	subkey3value = sorted_number_occurences[i]
	i += 1
	for original_key in d:
		if original_key in sort_by_dict.keys() and sort_by_dict[original_key] == subkey3value:
			# Just adding faux dict {original_key: d[original_key]} to a results list
			results_list.append({original_key: d[original_key]})

print results_list