**π‘ Problem Formulation:** In Python programming, a common task is to determine how often a specific sublist appears within a larger list. This can be useful in various applications such as data analysis, pattern recognition, and algorithm development. For example, if our input is `[1, 2, 1, 2, 3, 4, 1, 2]`

and the sublist we are looking for is `[1, 2]`

, the desired output would be `2`

as the sublist [1, 2] appears twice in the given list.

## Method 1: Using a for-loop and slicing

This method involves iterating through the parent list and using list slicing to match the sublist. Itβs a straightforward approach which doesn’t require any additional modules and is thus easily readable and implementable.

Here’s an example:

def count_sublist_frequency(lst, sublist): count = 0 sub_len = len(sublist) for i in range(len(lst) - sub_len + 1): if lst[i:i + sub_len] == sublist: count += 1 return count # Example usage main_list = [1, 2, 1, 2, 3, 4, 1, 2] sub_list = [1, 2] print(count_sublist_frequency(main_list, sub_list))

**Output:** 2

In the provided code snippet, the function `count_sublist_frequency()`

accepts the main list and the sublist to search for. It initializes a counter to zero, then iteratively checks slices of the main list, each the length of the sublist, incrementing the counter every time a match is found. The final result is the count of the frequency at which the sublist occurs.

## Method 2: Using Collections Module

The `collections`

module provides specialized container datatypes. In this case, `Collections.Counter()`

can be employed to count hashable objects. It can be used to count slices of the list as well, helping determine the frequency of sublists.

Here’s an example:

from collections import Counter def count_sublist_frequency(lst, sublist): sub_len = len(sublist) slices = [tuple(lst[i:i + sub_len]) for i in range(len(lst) - sub_len + 1)] return Counter(slices)[tuple(sublist)] # Example usage main_list = [1, 2, 1, 2, 3, 4, 1, 2] sub_list = [1, 2] print(count_sublist_frequency(main_list, sub_list))

**Output:** 2

The function `count_sublist_frequency()`

creates all possible slices of the list that are the same length as the sublist, converting them to tuples which are hashable and hence can be counted using `Counter`

. It then directly accesses the count for the given tuple equivalent of the sublist using the generated counter dictionary.

## Method 3: Using itertools’ groupby for consecutive sublists

Python’s `itertools.groupby()`

function can be handy when counting consecutive sublists. This approach is efficient for finding repeating patterns directly after each other in a sequence.

Here’s an example:

from itertools import groupby def count_consecutive_sublist(lst, sublist): count = 0 consecutive_groups = (len(list(group)) for _, group in groupby(lst)) return sum(sub_len == sublist_length for sub_len in consecutive_groups) # Example usage main_list = [1, 1, 2, 2, 2, 1, 2, 1, 1, 2, 2] sub_list = [1, 2] print(count_consecutive_sublist(main_list, sub_list))

**Output:** 3

This approach differs slightly, as it looks specifically for consecutive occurrences of elements of the sublist within the main list. The function `count_consecutive_sublist()`

uses `groupby()`

from the `itertools`

module to group repeating elements together, counts the length of each group, and then sums the instances where this length matches the length of the sublist.

## Method 4: Using list comprehension and sum function

This Pythonic way leverages the succinctness of list comprehension combined with the `sum()`

function to count the frequency of a sublist. It’s a more condensed and often faster technique for smaller lists.

Here’s an example:

def count_sublist_frequency(lst, sublist): sub_len = len(sublist) return sum(lst[i:i + sub_len] == sublist for i in range(len(lst) - sub_len + 1)) # Example usage main_list = [1, 2, 1, 2, 3, 4, 1, 2] sub_list = [1, 2] print(count_sublist_frequency(main_list, sub_list))

**Output:** 2

This streamlined code snippet defines a single-line function `count_sublist_frequency()`

that uses a list comprehension to create a list of boolean values that represent whether a slice of the main list matches the sublist, then simply sums these boolean values to get the total count of matches.

## Bonus One-Liner Method 5: Using a generator expression

For those who enjoy concise, functional programming style, a generator expression provides an elegant alternative. A generator expression for this problem reduces memory usage, as it avoids the creation of an intermediate list.

Here’s an example:

count_sublist_frequency = lambda lst, sublist: sum(1 for i in range(len(lst) - len(sublist) + 1) if lst[i:i+len(sublist)] == sublist) # Example usage main_list = [1, 2, 1, 2, 3, 4, 1, 2] sub_list = [1, 2] print(count_sublist_frequency(main_list, sub_list))

**Output:** 2

This example features a lambda function that uses a generator expression, iterating over indices and yielding a count of 1 each time a matching slice is found. The `sum()`

function then iterates over this generator to add up the counts, providing the total frequency of the sublist.

## Summary/Discussion

**Method 1: For-loop and Slicing.**Simple and clear. No external dependencies. May be less efficient with very large lists.**Method 2: Collections Module.**Utilizes powerful standard library modules. Efficient for repeated sublist searches. Slightly more complex.**Method 3: itertools.groupby() for Consecutive Sublists.**Best for counting consecutive patterns. Less flexible for non-consecutive sublists.**Method 4: List Comprehension and Sum.**Pythonic and concise. Efficient for smaller lists. Readability may suffer for those unfamiliar with the style.**Method 5: Generator Expression.**Memory efficient. Functional style that is very concise. Could be less readable for beginners.