Question

I have a sequence like [0, 1, 0, 1, 0, 1, 0] and I need a function to remove repeated adjacent sequence pairs, keeping the first one, and return [0, 1, 0]. These are some results I expect.

>>> remove_repeated_pairs([0, 1])
[0, 1]
>>> remove_repeated_pairs([0, 1, 0])
[0, 1, 0]
>>> remove_repeated_pairs([0, 1, 0, 1])
[0, 1]
>>> remove_repeated_pairs([0, 1, 0, 1, 0])
[0, 1, 0]
>>> remove_repeated_pairs([2, 0, 1, 0, 1, 0])
[2, 0, 1, 0]
>>> remove_repeated_pairs([1, 2, 0, 1, 0, 1, 0])
[1, 2, 0, 1, 0]

first edition:

I tried this code:

def remove_repeated_pairs(seq):
    result = []
    for i in range(0, len(seq), 2):
        if len(result) >= 2:
            last_seq = result[-2:]
        else:
            last_seq = None
        pair = seq[i:i + 2]
        if pair != last_seq:
            result.extend(pair)
    return result

But it doesn’t works with this:

>>> remove_repeated_pairs([1, 3, 0, 2, 1, 2, 1, 3, 0])
[1, 3, 0, 2, 1, 2, 1, 3, 0]

The right answer should be [1, 3, 0, 2, 1, 3, 0]

Answer 1

I think that the issue comes from the fact that you go over the elements of your list 2 by 2 (for i in range(0, len(seq), 2).
So if a repeated pair starts on an odd place, you won’t detect it – as in the last example you give.

I would try something like:

def remove_repeated_pairs(l):
    i = 2;
    while i < len(l)-1:
            if l[i] == l[i-2] and l[i+1]==l[i-1]:
                l.pop(i);
                l.pop(i);
            else:
                i+=1;

    return l;

Regards,