**π‘ Problem Formulation:** In many algorithmic problems, it is required to find the count of consecutive identical elements in a list or a string. For instance, given the input `["a", "a", "b", "b", "b", "a"]`

, the desired output is a count of consecutive repetition which would be `[2, 3, 1]`

, indicating that “a” repeats 2 times, then “b” repeats 3 times, followed by “a” with 1 occurrence.

## Method 1: Using Itertools Groupby

This method includes using the `groupby()`

function from Python’s `itertools`

module to group the consecutive identical elements, and then just count the size of each group. It’s efficient and takes advantage of Python’s built-in functionality for handling iterables.

Here’s an example:

from itertools import groupby def consecutive_elem_count(lst): return [sum(1 for _ in group) for _, group in groupby(lst)] example_list = ["a", "a", "b", "b", "b", "a"] print(consecutive_elem_count(example_list))

Output:

[2, 3, 1]

This snippet defines a function `consecutive_elem_count()`

which processes a list. It utilizes `groupby()`

to group consecutive identical elements and computes their count using a generator expression within a list comprehension.

## Method 2: Naive Loop Approach

The naive loop approach involves iterating over the list of elements, keeping track of consecutive repeats, and updating the count whenever a change in element is observed. This approach is straightforward and doesn’t require any special Python modules.

Here’s an example:

def consecutive_elem_count(lst): if not lst: return [] count_list, count = [], 1 for i in range(1, len(lst)): if lst[i] == lst[i-1]: count += 1 else: count_list.append(count) count = 1 count_list.append(count) return count_list example_list = ["a", "a", "b", "b", "b", "a"] print(consecutive_elem_count(example_list))

Output:

[2, 3, 1]

The code defines a function `consecutive_elem_count()`

which goes through the list using a for loop, tracking consecutive repeats and storing their counts to a separate list, which is then returned.

## Method 3: Using Regular Expressions

Regular expressions can be used to find consecutive identical characters in a string by matching patterns. This method is powerful as regex offers a lot of flexibility and is very concise for pattern matching tasks.

Here’s an example:

import re def consecutive_elem_count(s): return [len(match) for match in re.findall(r'(.)\1*', s)] example_string = "aabbbba" print(consecutive_elem_count(example_string))

Output:

[2, 4, 1]

This snippet creates the `consecutive_elem_count()`

function that uses Regex’s `findall()`

function to detect groups of repeating characters in a string and counts them.

## Method 4: Using Numpy

Numpy, a powerful numerical processing library, can be used for counting consecutive elements by leveraging boolean masking and the `diff`

function, useful on numerical data.

Here’s an example:

import numpy as np def consecutive_elem_count(arr): diff_arr = np.diff(arr, prepend=np.nan) return np.diff(np.where(diff_arr != 0)[0], append=arr.size) example_array = np.array(["a", "a", "b", "b", "b", "a"]) print(consecutive_elem_count(example_array))

Output:

[2 3 1]

The function `consecutive_elem_count()`

computes the differences in consecutive array elements and then counts the occurrences before each change. This method exploits the efficiency of Numpy’s vectorized operations.

## Bonus One-Liner Method 5: List Comprehension with zip

Python’s list comprehension feature, combined with the `zip`

function, can help find consecutive elements count in a very succinct way.

Here’s an example:

lst = ["a", "a", "b", "b", "b", "a"] consecutive_counts = [sum(1 for _ in group) for group in (list(g) for k, g in zip(lst, lst[1:] + lst[:1]) if k == g)] print(consecutive_counts)

Output:

[2, 3, 1]

This one-liner uses list comprehension to iterate over paired elements and count consecutive matches. Note that it builds temporary lists to count iterations, making it less efficient.

## Summary/Discussion

**Method 1: Groupby from Itertools.**Efficient and Pythonic. Sticks with built-in libraries. Not as easy to read for beginners.**Method 2: Naive Loop Approach.**Straightforward, easy to understand. Can be slow for large lists. No third-party libraries required.**Method 3: Using Regular Expressions.**Powerful pattern matching. Less efficient for non-string data. Might be overkill for simple lists.**Method 4: Using Numpy.**Fast for numerical data. Overhead for importing a large library. Not suitable for small or single-use scripts.**Bonus Method 5: List Comprehension with zip.**One-liner, elegant. Less efficient due to temporary list construction. Might be tricky to understand at first glance.