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Python – Remove alternate consecutive duplicates
• Last Updated : 01 Aug, 2020

Given list of elements, remove alternate consecutive duplicates of elements.

Input : test_list = [5, 5, 5, 5, 6, 6]
Output : [5, 5, 6]
Explanation : Alternate occ. of 5 and 6 are removed.

Input : test_list = [5, 5, 5, 5]
Output : [5, 5]
Explanation : Alternate occ. of 5 are removed.

Method : Using loop + remove()

The combination of above functions can be used to solve this problem. In this, we iterate each element and use additional previous element variable to keep track of alternate criteria. The removal is performed using remove().

## Python3

 `# Python3 code to demonstrate working of ``# Remove alternate consecutive duplicates``# Using loop + remove()`` ` `# initializing lists``test_list ``=` `[``5``, ``5``, ``5``, ``5``, ``6``, ``6``, ``8``, ``3``, ``3``, ``8``]`` ` `# printing original list``print``(``"The original list : "` `+` `str``(test_list))`` ` `# Using loop to iterate through elements``# element to keep track``temp ``=` `test_list[``0``]``count ``=` `0``org_list ``=` `test_list``idx ``=` `0``while``(``1``):``     ` `    ``# break when idx greater than size``    ``if` `idx >``=` `len``(org_list):``        ``break``     ` `    ``# check for alternates ``    ``if` `count ``%` `2` `and` `temp ``=``=` `test_list[idx]:``        ``test_list.remove(test_list[idx])``        ``idx ``=` `idx ``-` `1``        ``count ``+``=` `1``        ``temp ``=` `test_list[idx]``    ``else``:``         ` `        ``# keeping track of alternate index increment``        ``# and assignment``        ``if` `temp !``=` `test_list[idx]:``            ``count ``=` `1``            ``temp ``=` `test_list[idx]``        ``else` `:``            ``count ``+``=` `1``    ``idx ``=` `idx ``+` `1`` ` `# printing result ``print``(``"List after alternate duplicates removal : "` `+` `str``(test_list))`
Output
```The original list : [5, 5, 5, 5, 6, 6, 8, 3, 3, 8]
List after alternate duplicates removal : [5, 5, 6, 8, 3, 8]
```

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