Python | Sort a Dictionary

Python, Given a dictionary, perform sort, basis on keys or values. [ applicable Python >=3.6v ].

Input : test_dict = {“Gfg” : 5, “is” : 7, “Best” : 2} Output : {‘Best’: 2, ‘Gfg’: 5, ‘is’: 7}, {‘is’: 7, ‘Gfg’: 5, ‘Best’: 2} Explanation : Sorted by keys, in ascending and reverse order. Input : test_dict = {“Best” : 2, “for” : 9, “geeks” : 8} Output : {‘Best’: 2, ‘Gfg’: 5, ‘for’: 9}, {‘for’: 9, ‘geeks’: 8, ‘Best’: 2} Explanation : Sorted by values, in ascending and reverse order.

Case 1 : Sort by Keys

This task is performed using sorted(), in this, we extract the keys using 1st index of items of dictionary extracted by items(), and pass it in key as custom lambda function to get sorted by keys. The “reverse=True” is added to perform reverse sort.

Python3

# Python3 code to demonstrate working of
# Sort a Dictionary
# Sort by Keys

# initializing dictionary
test_dict = {"Gfg" : 5, "is" : 7, "Best" : 2, "for" : 9, "geeks" : 8}

# printing original dictionary
print("The original dictionary is : " + str(test_dict))

# using items() to get all items
# lambda function is passed in key to perform sort by key
res = {key: val for key, val in sorted(test_dict.items(), key = lambda ele: ele[0])}

# printing result
print("Result dictionary sorted by keys : " + str(res))

# using items() to get all items
# lambda function is passed in key to perform sort by key
# adding "reversed = True" for reversed order
res = {key: val for key, val in sorted(test_dict.items(), key = lambda ele: ele[0], reverse = True)}

# printing result
print("Result dictionary sorted by keys ( in reversed order ) : " + str(res))

Output

The original dictionary is : {'Gfg': 5, 'is': 7, 'Best': 2, 'for': 9, 'geeks': 8}
Result dictionary sorted by keys : {'Best': 2, 'Gfg': 5, 'for': 9, 'geeks': 8, 'is': 7}
Result dictionary sorted by keys ( in reversed order ) : {'is': 7, 'geeks': 8, 'for': 9, 'Gfg': 5, 'Best': 2}

Case 2 : Sort by Values 

This task can be performed in similar way as above, the only difference being for extracting values, 2nd element of items() is passed as comparator. 

Python3

# Python3 code to demonstrate working of
# Sort a Dictionary
# Sort by Values

# initializing dictionary
test_dict = {"Gfg" : 5, "is" : 7, "Best" : 2, "for" : 9, "geeks" : 8}

# printing original dictionary
print("The original dictionary is : " + str(test_dict))

# using items() to get all items
# lambda function is passed in key to perform sort by key
# passing 2nd element of items()
res = {key: val for key, val in sorted(test_dict.items(), key = lambda ele: ele[1])}

# printing result
print("Result dictionary sorted by values : " + str(res))

# using items() to get all items
# lambda function is passed in key to perform sort by key
# passing 2nd element of items()
# adding "reversed = True" for reversed order
res = {key: val for key, val in sorted(test_dict.items(), key = lambda ele: ele[1], reverse = True)}

# printing result
print("Result dictionary sorted by values ( in reversed order ) : " + str(res))

Output

The original dictionary is : {'Gfg': 5, 'is': 7, 'Best': 2, 'for': 9, 'geeks': 8}
Result dictionary sorted by values : {'Best': 2, 'Gfg': 5, 'is': 7, 'geeks': 8, 'for': 9}
Result dictionary sorted by values ( in reversed order ) : {'for': 9, 'geeks': 8, 'is': 7, 'Gfg': 5, 'Best': 2}

Method#3:Using collections.OrderedDict() and sorted() 

Approach

this approach uses the sorted() function to sort a dictionary by its values in either ascending or descending order. The sorted() function is called with the items() method of the dictionary and a key function that returns the second element of each tuple (i.e., the values) or their negation. The resulting list of tuples is passed to the OrderedDict() constructor to create a new ordered dictionary with the same key-value pairs as the original dictionary but sorted by value.

Algorithm

1. Call the sorted() function on the dictionary ‘test_dict’, passing a lambda function as the ‘key’ argument.
2. The lambda function takes each key-value pair as input and returns the key or value to sort by, depending on the desired order.
3. Use the sorted() function to return a list of sorted key-value pairs.
4. Pass the sorted list to the OrderedDict() constructor to create a new ordered dictionary.
5. Return the ordered dictionary.

Python3

from collections import OrderedDict
from operator import itemgetter

def sort_dict_by_value(test_dict):
    sorted_list = sorted(test_dict.items(), key=itemgetter(1))
    return OrderedDict(sorted_list)

def sort_dict_by_value_reverse(test_dict):
    sorted_list = sorted(test_dict.items(), key=itemgetter(1), reverse=True)
    return OrderedDict(sorted_list)

test_dict = {"Gfg" : 5, "is" : 7, "Best" : 2, "for" : 9, "geeks" : 8}
print(sort_dict_by_value(test_dict))
print(sort_dict_by_value_reverse(test_dict))

Output

OrderedDict([('Best', 2), ('Gfg', 5), ('is', 7), ('geeks', 8), ('for', 9)])
OrderedDict([('for', 9), ('geeks', 8), ('is', 7), ('Gfg', 5), ('Best', 2)])

Time Complexity: O(N log N), where N is the number of key-value pairs in the dictionary.
Space Complexity: O(N), as we are creating a new ordered dictionary to store the sorted key-value pairs.

Method 4 :  use the sorted() method with a lambda function as the key parameter. 

Here are the steps:

1. Define the dictionary to be sorted.
2. Use the sorted() method to sort the dictionary by values.
3. Pass a lambda function as the key parameter to the sorted() method to specify that the sorting should be done by values.
4. Use the dict() constructor to create a new dictionary from the sorted list of tuples.

Python3

def sort_dict_by_value_lambda(test_dict):
    sorted_list = sorted(test_dict.items(), key=lambda x: x[1])
    return dict(sorted_list)

def sort_dict_by_value_lambda_reverse(test_dict):
    sorted_list = sorted(test_dict.items(), key=lambda x: x[1], reverse=True)
    return dict(sorted_list)

test_dict = {"Gfg" : 5, "is" : 7, "Best" : 2, "for" : 9, "geeks" : 8}
print(sort_dict_by_value_lambda(test_dict))
print(sort_dict_by_value_lambda_reverse(test_dict))

Output

{'Best': 2, 'Gfg': 5, 'is': 7, 'geeks': 8, 'for': 9}
{'for': 9, 'geeks': 8, 'is': 7, 'Gfg': 5, 'Best': 2}

Time complexity: O(n log n) where n is the number of items in the dictionary.

Auxiliary space: O(n) to store the sorted list of tuples. The dict() constructor takes O(n) time to create a new dictionary from the sorted list.