>>> # regular unsorted dictionary
>>> d = {'banana': 3, 'apple': 4, 'pear': 1, 'orange': 2}

>>> # dictionary sorted by key
>>> 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)])

most_common([n])¶
Return a list of the n most common elements and their counts from the most common to the least. If n is omitted or None, most_common() returns all elements in the counter.
Elements with equal counts are ordered arbitrarily:

>>> Counter('abracadabra').most_common(3)
[('a', 5), ('r', 2), ('b', 2)]

#Update: As of Python 3.7 (and CPython 3.6), standard dict is 
	#guaranteed to preserve order and is more performant than OrderedDict. 
#(For backward compatibility and especially readability, 
	#however, you may wish to continue using OrderedDict.)

>>> keywords = ['foo', 'bar', 'bar', 'foo', 'baz', 'foo']

>>> list(dict.fromkeys(keywords))
['foo', 'bar', 'baz']

# A Python program to demonstrate working of OrderedDict
from collections import OrderedDict
 
print("This is a Dict:\n")
d = {}
d['a'] = 1
d['b'] = 2
d['c'] = 3
d['d'] = 4
 
for key, value in d.items():
    print(key, value)
 
print("\nThis is an Ordered Dict:\n")
od = OrderedDict()
od['a'] = 1
od['b'] = 2
od['c'] = 3
od['d'] = 4
 
for key, value in od.items():
    print(key, value)