Select sublists from python list, beginning and ending on the same element

Question

I have a (very large) list similar to:

a = ['A', 'B', 'A', 'B', 'A', 'C', 'D', 'E', 'D', 'E', 'D', 'F', 'G', 'A', 'B']

and I want to extract from it a list of lists like:

result = [['A', 'B', 'A', 'B', 'A'], ['D', 'E', 'D', 'E', 'D']]

The repeating patterns can be different, for example there can also be intervals such as:

['A', 'B', 'C', 'A', 'D', 'E', 'A'] (with a 'jump' over two elements)

I have written a very simple code that seems to work:

tolerance = 2
counter = 0
start, stop = 0, 0
for idx in range(len(a) - 1):
    if a[idx] == a[idx+1] and counter == 0:
        start = idx
        counter += 1
    elif a[idx] == a[idx+1] and counter != 0:
        if tolerance <= 0: 
            stop = idx
        tolerance = 2
    elif a[idx] != a[idx+1]:
        tolerance -= 1
    if start != 0 and stop != 0:
        result = [a[start::stop]]

But 1) it is very cumbersome, 2) I need to apply this to very large lists, so is there a more concise and faster way of implementing it?

EDIT: As @Kasramvd correctly pointed out, I need the largest set that satisfies the requirement of (at most a tolerance number of jumps between the start and end elements), so I take:

['A', 'B', 'A', 'B', 'A'] instead of [ 'B', 'A', 'B' ]

because the former includes the latter.

Also it would be good if the code can select elements UP TO the certain tolerance, for example if the tolerance (maximum number of elements not equal to the start or end element) is 2, it should also return sets as:

['A', 'A', 'A', 'B', 'A', 'B', 'A', 'C', 'D', 'A']

with tolerances 0, 1 and 2.

Answer 1

Solution without any extra copying of lists other than the sublist results:

def sublists(a, tolerance):
    result = []
    index = 0

    while index < len(a):
        curr = a[index]

        for i in range(index, len(a)):
            if a[i] == curr:
                end = i
            elif i - end > tolerance:
                break

        if index != end:
            result.append(a[index:end+1])
        index += end - index + 1

    return result

Usage is simply as follows:

a = ['A', 'B', 'A', 'B', 'A', 'C', 'D', 'E', 'D', 'E', 'D', 'F', 'G', 'A', 'B']

sublists(a, 0)  # []
sublists(a, 1)  # [['A', 'B', 'A', 'B', 'A'], ['D', 'E', 'D', 'E', 'D']]
sublists(a, 2)  # [['A', 'B', 'A', 'B', 'A'], ['D', 'E', 'D', 'E', 'D']]

Possible solution to extra requirement as specified in the comments:

if i > index and a[i] == a[i-1] == curr:
    end = i - 1
    break
elif a[i] == curr:
    end = i
elif i - end > tolerance:
    break

Note: I've not tested this thoroughly.