[Tutor] removing consecutive duplicates from list

[dn]

On 21/04/2021 05.27, Peter Otten wrote:
> On 20/04/2021 18:48, Manprit Singh wrote:
> 
>> Consider a list given below:
>> lst = [2, 3, 3, 4, 5, 5, 3, 7, 9, 9, 4]
>> i need to remove consecutive duplicates from list lst:
>> the answer must be :
>>
>> [2, 3, 4, 5, 3, 7, 9, 4]
>>
>> The code that i have written to solve it, is written below:
>> lst = [2, 3, 3, 4, 5, 5, 3, 7, 9, 9, 4]
>> ls = lst[1:]+[object()]
>> [x for x, y in zip(lst, ls) if x != y]
>>
>> The list comprehension gives the desired result. just need to know if
>> this
>>
>> program can be done in a more readable and less complex way.


> The itertools module has many tools that you can use to deal with
> problems like the above:
...

>>>> from itertools import groupby
>>>> wanted == [k for k, g in groupby(lst)]
> True
> 
> Pick your favourite ;)


Whilst, admiring your (@Manprit) one-liner and @Peter's grasp of the
itertools library; am imagining yellow, if not red, flags.

Taking the term "less complex", let's ask: how many programmers
(including yourself in six months' time) will be able to scan any of
these and conclude?deduce that their objective is the removal of
consecutive duplicates?

At the very least, please enclose your 'brilliance' within a function,
which will give it a (better than a comment) label/name, eg:

def remove_consecutive_duplicates( source ):
    return [k for k, g in groupby(lst)]


Ultimately, I favor @Alan's "explicit form" approach (shhhh! Don't tell
him, else he'll fall off his chair...).

Here's me dusting-off an old solution to this problem, which we used to
give ComSc students - updated to (I think) v3.9+:



from typing import Iterable

lst = [2, 3, 3, 4, 5, 5, 3, 7, 9, 9, 4]
result = [2, 3, 4, 5, 3, 7, 9, 4]

def remove_consecutive_duplicates( source:Iterable )->list:
    """Clean input iterable (containing any data-types),
       removing consecutive duplicates.
    """
    cleaned = list()
    current_value = None
    for this_element in source:
        if this_element != current_value:
            cleaned.append( this_element )
            current_value = this_element
    return cleaned

print( "Correct?", result == remove_consecutive_duplicates( lst ) )


Plus:

remove_consecutive_duplicates( "abba" ) == ['a', 'b', 'a']
-- 
Regards,
=dn