5 Best Ways to Sort Python Dictionaries by Size

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💡 Problem Formulation: When dealing with Python dictionaries, it is often useful to order them based on the size of their values. Whether these values are lists, sets, or another collection data type, being able to sort dictionaries by the length of their contained elements is a common task. For instance, given a dictionary {'a': [1, 2, 3], 'b':[1, 2], 'c':[1]}, one might want to sort the dictionary to get [('c', [1]), ('b', [1, 2]), ('a', [1, 2, 3])] as the output, sorted from the smallest to the largest value size.

Method 1: Using the sorted() Function with a Lambda

The sorted() built-in function is versatile and can be used with a lambda function as the key argument to sort a dictionary by the size of its values. This method is straightforward and utilizes the power of lambda functions to define a custom sort order quickly and concisely.

Here’s an example:

d = {'apple': [3, 2, 1], 'banana': [1], 'cherry': [2, 3, 1, 4]}
sorted_d = sorted(d.items(), key=lambda item: len(item[1]))
print(sorted_d)

Output:

[('banana', [1]), ('apple', [3, 2, 1]), ('cherry', [2, 3, 1, 4])]

This code snippet takes a dictionary with fruit names as keys and lists of numbers as values. It sorts the dictionary items by the length of the lists using a lambda function as the sorting key. The items() method is used to get a view of the dictionary’s items, which is then passed to sorted(), which performs the sorting based on the length of the second element (the list) of each tuple in the items view.

Method 2: Sorting with an Explicit Function

For more complexity or when lambda functions are not expressive enough, one can define an explicit function to pass as the key argument to sorted(). This approach offers more clarity and potentially more complex sorting logic.

Here’s an example:

def sort_by_value_size(item):
    return len(item[1])

d = {'apple': [3, 2, 1], 'banana': [1], 'cherry': [2, 3, 1, 4]}
sorted_d = sorted(d.items(), key=sort_by_value_size)
print(sorted_d)

Output:

[('banana', [1]), ('apple', [3, 2, 1]), ('cherry', [2, 3, 1, 4])]

This code defines a custom sorting function sort_by_value_size() that takes a dictionary item and returns the size of the item’s value. It’s used as the key function for sorting the dictionary items. This approach is similar to Method 1 but may be preferred for enhanced readability or when dealing with more complex sorting logic than what a lambda can comfortably represent.

Method 3: Using operator.itemgetter()

The operator module provides a way to retrieve the item of interest directly, making the sort more explicit and potentially more efficient. The itemgetter() function can be used as the key function in the sorted() call, which can result in slightly better performance.

Here’s an example:

from operator import itemgetter

d = {'apple': [3, 2, 1], 'banana': [1], 'cherry': [2, 3, 1, 4]}
sorted_d = sorted(d.items(), key=lambda item: len(itemgetter(1)(item)))
print(sorted_d)

Output:

[('banana', [1]), ('apple', [3, 2, 1]), ('cherry', [2, 3, 1, 4])]

In this snippet, we import itemgetter() from the operator module to retrieve the second element of the dictionary item. The sorted() function sorts the items by the length of these second elements, precisely as before. This method can result in a performance benefit, especially for large data sets, although the readability may not be as clear as the simple lambda approach.

Method 4: Sorting In-Place Using dict Methods

Sometimes, we might want to sort a dictionary in-place. While dictionaries before Python 3.7 were not officially ordered, the language has since then guaranteed the order of items. Thus, methods like update() could be used in a creative way to reorder a dictionary.

Here’s an example:

d = {'apple': [3, 2, 1], 'banana': [1], 'cherry': [2, 3, 1, 4]}
d.update(sorted(d.items(), key=lambda item: len(item[1])))
print(d)

Output:

{'banana': [1], 'apple': [3, 2, 1], 'cherry': [2, 3, 1, 4]}

This code snippet sorts a dictionary in-place by using the update() method on the original dictionary with the sorted items. It’s worth noting that a dictionary can only be reliably sorted in-place if using Python 3.7 or later. This method retains the semantics of changing the original dictionary instead of creating a new one.

Bonus One-Liner Method 5: Sorting Using Comprehensions

Python’s comprehensions are a concise way to achieve tasks that would otherwise need more verbose loops. Dictionary comprehensions can sort a dictionary by the size of its values in a single line of code.

Here’s an example:

d = {'apple': [3, 2, 1], 'banana': [1], 'cherry': [2, 3, 1, 4]}
sorted_d = {k: v for k, v in sorted(d.items(), key=lambda item: len(item[1]))}
print(sorted_d)

Output:

{'banana': [1], 'apple': [3, 2, 1], 'cherry': [2, 3, 1, 4]}

This one-liner uses a dictionary comprehension to create a new dictionary that’s sorted by the length of its values. The sorted() function with a lambda key provides the sorted order. This is an elegant and Pythonic way of accomplishing the task, but it does create a new dictionary as the result.

Summary/Discussion

  • Method 1: Lambda with sorted(). Quick and concise. Less readable with complex logic.
  • Method 2: Explicit function with sorted(). Clearer logic. Slightly more verbose.
  • Method 3: operator.itemgetter() with sorted(). More explicit and efficient. Can be less readable.
  • Method 4: In-place sorting with update(). Alter the original dictionary. Only works reliably in Python 3.7 and later.
  • Method 5: Dictionary comprehension. Elegant and Pythonic way to create a sorted dictionary. Generates a new dictionary.