**π‘ Problem Formulation:** This article addresses the issue of identifying and retrieving duplicate sets from a collection of sets in Python. For instance, given a list of sets like `[{1, 2}, {3, 4}, {1, 2}]`

, the desired output would be `[{1, 2}]`

as it appears more than once.

## Method 1: Using a Dictionary for Frequency Count

This method entails creating a dictionary to keep track of the frequency of each set. Due to the immutability requirement of dictionary keys, the sets must first be converted to frozensets. The program iterates over the list, counting each occurrence and then extracting the sets that appear more than once.

Here’s an example:

def find_duplicate_sets(sets_list): freq_dict = {} for s in sets_list: frozen_s = frozenset(s) freq_dict[frozen_s] = freq_dict.get(frozen_s, 0) + 1 return [set(item) for item, count in freq_dict.items() if count > 1] # Example usage sets_list = [{1, 2}, {3, 4}, {1, 2}] duplicates = find_duplicate_sets(sets_list) print(duplicates)

Output:

[{1, 2}]

The function `find_duplicate_sets()`

converts each set to a frozenset and then increments its count in the dictionary. Duplicates are identified by checking for a count higher than one and then converted back to sets for the output.

## Method 2: Using Collections.Counter

This strategy utilizes Python’s `collections.Counter`

class, which is a dictionary subclass for counting hashable objects. Sets are not directly hashable, so they are converted to frozensets before passing them to the Counter.

Here’s an example:

from collections import Counter def find_duplicate_sets(sets_list): counts = Counter(frozenset(s) for s in sets_list) return [set(item) for item, count in counts.items() if count > 1] # Example usage sets_list = [{1, 2}, {1, 2}, {3, 4}] duplicates = find_duplicate_sets(sets_list) print(duplicates)

Output:

[{1, 2}]

The `find_duplicate_sets()`

function creates a Counter object that automatically counts the frequency of each frozenset and then extracts duplicates based on a count greater than one, returning the original sets.

## Method 3: Sorting and Grouping

Sorting the list of sets and then grouping them together can help identify duplicates. This method uses `itertools.groupby`

, which groups consecutive elements. The list of sets needs to be sorted based on a criteria that allows grouping, such as sorted tuples from the sets.

Here’s an example:

from itertools import groupby def find_duplicate_sets(sets_list): sets_list.sort(key=lambda s: sorted(s)) return [set(next(group)) for key, group in groupby(sets_list, sorted) if len(list(group)) > 1] # Example usage sets_list = [{1, 2}, {2, 1}, {3, 4}] duplicates = find_duplicate_sets(sets_list) print(duplicates)

Output:

[{1, 2}]

The function `find_duplicate_sets()`

sorts the list and then applies `groupby()`

. The generator `group`

has more than one item if there are duplicates, which are then returned as sets.

## Method 4: Using Set Operations

With set operations, one can identify duplicates by adding sets to a new list and checking if they’re already present, exploiting the property that sets can’t have duplicate elements.

Here’s an example:

def find_duplicate_sets(sets_list): unique_sets = set() duplicates = set() for s in sets_list: frozen_s = frozenset(s) if frozen_s in unique_sets: duplicates.add(frozen_s) else: unique_sets.add(frozen_s) return [set(s) for s in duplicates] # Example usage sets_list = [{1, 2}, {1, 2}, {2, 3}] duplicates = find_duplicate_sets(sets_list) print(duplicates)

Output:

[{1, 2}]

The function `find_duplicate_sets()`

iterates over the list, converting each set into a frozenset and checking if it’s in the `unique_sets`

. If so, it’s a duplicate and added to the `duplicates`

set.

## Bonus One-Liner Method 5: Using List Comprehension and Collections.Counter

This one-liner method combines list comprehension with `Counter`

to find duplicate sets efficiently in a single line of code.

Here’s an example:

from collections import Counter sets_list = [{1, 2}, {3, 4}, {1, 2}] duplicates = [set(item) for item, count in Counter(frozenset(s) for s in sets_list).items() if count > 1] print(duplicates)

Output:

[{1, 2}]

This one-liner uses a list comprehension to construct a list of duplicates by filtering out counts greater than one from a `Counter`

that’s populated with frozensets derived from the original list of sets.

## Summary/Discussion

**Method 1: Dictionary for Frequency Count.**Simple and explicit method. May not be the most efficient due to the manual iteration and dictionary updating involved.**Method 2: Collections.Counter.**Concise and relies on standard library convenience. However, lacks the simplicity of a one-liner solution.**Method 3: Sorting and Grouping.**Elegant and useful for sorted output. Requires the list to be mutable and may not be as efficient due to sorting.**Method 4: Using Set Operations.**Straightforward logic using properties of sets. Operationally simple but requires additional space for tracking unique sets and duplicates.**Bonus Method 5: One-Liner List Comprehension.**Compact and Pythonic. Ideal for short scripts or inline usage. However, this method might be less readable for beginners.