Python Program to Sort Dictionary by Value

You can sort a dictionary by value in Python using sorted(your_dict.items(), key=lambda item: item[1]) which returns a list of tuples sorted by the dictionary's values.

Examples

Input{'a': 3, 'b': 1, 'c': 2}

Output[('b', 1), ('c', 2), ('a', 3)]

Input{'apple': 10, 'banana': 5, 'cherry': 7}

Output[('banana', 5), ('cherry', 7), ('apple', 10)]

Input{}

Output[]

How to Think About It

To sort a dictionary by its values, think of the dictionary as a list of pairs (key, value). You want to arrange these pairs based on the second part (the value). Using sorted() with a function that picks the value part helps you do this easily.

Algorithm

1

Take the dictionary as input.

2

Convert the dictionary into a list of (key, value) pairs.

3

Use a sorting function that compares pairs by their value part.

4

Return the sorted list of pairs.

Code

python

my_dict = {'a': 3, 'b': 1, 'c': 2}
sorted_items = sorted(my_dict.items(), key=lambda item: item[1])
print(sorted_items)

Output

[('b', 1), ('c', 2), ('a', 3)]

Dry Run

Let's trace sorting {'a': 3, 'b': 1, 'c': 2} by value through the code

1

Original dictionary

{'a': 3, 'b': 1, 'c': 2}

2

Convert to list of items

[('a', 3), ('b', 1), ('c', 2)]

3

Sort by second element (value)

Sorted list: [('b', 1), ('c', 2), ('a', 3)]

Key	Value
a	3
b	1
c	2

Why This Works

Step 1: Dictionary items as pairs

Using .items() converts the dictionary into pairs of (key, value) which can be sorted.

Step 2: Sorting with a key function

The key=lambda item: item[1] tells Python to sort by the value part of each pair.

Step 3: Result is a sorted list

The output is a list of tuples sorted by values, not a dictionary.

Alternative Approaches

Using operator.itemgetter

python

from operator import itemgetter
my_dict = {'a': 3, 'b': 1, 'c': 2}
sorted_items = sorted(my_dict.items(), key=itemgetter(1))
print(sorted_items)

This uses a built-in function for better readability and slightly faster sorting.

Sorting dictionary and creating a new dict

python

my_dict = {'a': 3, 'b': 1, 'c': 2}
sorted_dict = dict(sorted(my_dict.items(), key=lambda item: item[1]))
print(sorted_dict)

This returns a new dictionary sorted by values but note that dictionaries preserve order only in Python 3.7+.

Sorting in reverse order

python

my_dict = {'a': 3, 'b': 1, 'c': 2}
sorted_items = sorted(my_dict.items(), key=lambda item: item[1], reverse=True)
print(sorted_items)

Sorts the dictionary by values from highest to lowest.

Complexity: O(n log n) time, O(n) space

Time Complexity

Sorting the dictionary items takes O(n log n) time because sorting algorithms compare elements multiple times.

Space Complexity

The sorted function creates a new list of items, so it uses O(n) extra space.

Which Approach is Fastest?

Using operator.itemgetter is slightly faster than a lambda but both have the same complexity.

Approach	Time	Space	Best For
Lambda with sorted()	O(n log n)	O(n)	Simple and clear sorting by value
operator.itemgetter	O(n log n)	O(n)	Slightly faster and more readable
Creating new dict from sorted items	O(n log n)	O(n)	When you want a sorted dictionary (Python 3.7+)