How Can We Efficiently Remove Consecutive Duplicates from a List While Preserving Unique Elements in Python?

Eliminating Consecutive Duplicates and Preserving Distinct Elements

When working with lists, it's often necessary to eliminate consecutive duplicate elements. The question arises: how can we efficiently remove such elements while preserving distinct elements?

One approach is to iterate through the list, comparing each element with its successor. If they're identical, we delete the current element. This method, while functional, is relatively inefficient and lacks Pythonic elegance.

Using GroupBy for Elegant Removal of Duplicates

The Python itertools module provides a powerful function called groupby that can efficiently group consecutive elements. By utilizing groupby, we can achieve our goal with fewer lines of code and increased clarity.

For the first part of the question (finding unique elements), we can use groupby to group consecutive elements, then select the keys (which represent unique elements).

from itertools import groupby

L = [1,1,1,1,1,1,2,3,4,4,5,1,2]
unique_elements = [key for key, _group in groupby(L)]

Output:

[1, 2, 3, 4, 5, 1, 2]

Preserving Distinct Elements Using Sum

To further refine the solution, we need to eliminate elements that have consecutive duplicates. Again, groupby comes to our aid. By checking the length of each group, we can determine whether an element is distinct or not.

distinct_elements = [k for k, g in groupby(L) if len(list(g)) < 2]

Output:

[2, 3, 5, 1, 2]

Alternatively, we can use a generator expression to avoid creating a temporary list for each group and instead calculate the sum of the generator:

distinct_elements = [k for k, g in groupby(L) if sum(1 for i in g) < 2]

This provides a more compact and efficient implementation of the solution, effectively eliminating consecutive duplicates while preserving distinct elements.

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