How to Remove Adjacent Duplicates in a Python List – 5 Simple Methods

Introduction

Hi everyone, I’m Pipí. In a recent Python group chat a member asked how to process a list to remove adjacent duplicate values. The original list is

origin_lst = [0, 0, 1, 2, 3, 4, 4, 5, 6, 6, 6, 7, 8, 9, 4, 4]

and the desired output is [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 4].

Method 1 – itertools.groupby

Using itertools.groupby to collapse consecutive duplicates:

import itertools
origin_lst = [0, 0, 1, 2, 3, 4, 4, 5, 6, 6, 6, 7, 8, 9, 4, 4]
final_lst = [x[0] for x in itertools.groupby(origin_lst)]
print(final_lst)

The result matches the expectation.

Method 2 – List Comprehension

A list‑comprehension that checks each element against its predecessor:

origin_lst = [0, 0, 1, 2, 3, 4, 4, 5, 6, 6, 6, 7, 8, 9, 4, 4]
res = [origin_lst[i] for i in range(len(origin_lst)) if i == 0 or origin_lst[i] != origin_lst[i - 1]]
print(res)

Method 3 – Simple Loop

A straightforward loop that appends a value only when it differs from the previous one:

origin_lst = [0, 0, 1, 2, 3, 4, 4, 5, 6, 6, 6, 7, 8, 9, 4, 4]
result = [origin_lst[0]]
for i in range(1, len(origin_lst)):
    if origin_lst[i] != origin_lst[i-1]:
        result.append(origin_lst[i])
print(result)

Method 4 – Generator Function

A generator that yields values only when they differ from the previously yielded item:

origin_lst = [0, 0, 1, 2, 3, 4, 4, 5, 6, 6, 6, 7, 8, 9, 4, 4]

def del_adjacent(iterable):
    prev = object()
    for item in iterable:
        if item != prev:
            prev = item
            yield item

result = list(del_adjacent(origin_lst))
print(result)

Method 5 – enumerate

Using enumerate to compare each element with the previous one by index:

origin_lst = [0, 0, 1, 2, 3, 4, 4, 5, 6, 6, 6, 7, 8, 9, 4, 4]

lst_final = []
for index, val in enumerate(origin_lst):
    if val != origin_lst[index - 1]:
        lst_final.append(val)
print(lst_final)

Conclusion

The article presented a Python data‑processing challenge and offered multiple clear solutions, helping the community solve the problem efficiently. Thanks to the contributors who shared their ideas and code.