How can I delete all zeros except for x of them in every run of consecutive zeros within a list?

Question

For every run of x or more consecutive zeros in a list in Python, I would like to del all zeros in the run except for x of them. If x = 0, then delete all zeros.

I was thinking of a Python function that took a list, L, and a number, x, as inputs.

For example, let L = [7, 0, 12, 0, 0, 2, 0, 0, 0, 27, 10, 0, 0, 0, 0, 8].

• If x = 0, then return L = [7, 12, 2, 27, 10, 8]
• If x = 1, then return L = [7, 0, 12, 0, 2, 0, 27, 10, 0, 8]
• If x = 2, then return L = [7, 0, 12, 0, 0, 2, 0, 0, 27, 10, 0, 0, 8]
• If x = 3, then return L = [7, 0, 12, 0, 0, 2, 0, 0, 0, 27, 10, 0, 0, 0, 8]
• If x = 4, then return L = [7, 0, 12, 0, 0, 2, 0, 0, 0, 27, 10, 0, 0, 0, 0, 8] (Same as original L)
• If x >= 5, then return original L as there are no runs of 5 or more consecutive zeros.

Any help would be sincerely appreciated.

Answer 1

This is easy to do as a generator. Wrap your call to it in a list constructor if you want a fresh list with the zero-runs removed.

def compact_zero_runs(iterable, max_zeros):
    zeros = 0
    for i in iterable:
        if i == 0:
            zeros += 1
            if zeros <= max_zeros:
                yield i
        else:
            zeros = 0
            yield i

Answer 2

Using groupby:

def del_zeros(lst, n):
    lst = (list(j)[:n] if i else list(j) 
           for i,j in itertools.groupby(lst, key=lambda x:x==0))

    return [item for sublist in lst for item in sublist]

And the tests:

>>> [del_zeros(L, i) for i in range(5)]
[[7, 12, 2, 27, 10, 8],
 [7, 0, 12, 0, 2, 0, 27, 10, 0, 8],
 [7, 0, 12, 0, 0, 2, 0, 0, 27, 10, 0, 0, 8],
 [7, 0, 12, 0, 0, 2, 0, 0, 0, 27, 10, 0, 0, 0, 8],
 [7, 0, 12, 0, 0, 2, 0, 0, 0, 27, 10, 0, 0, 0, 0, 8]]

Answer 3

from itertools import groupby, chain, islice
from functools import partial
from operator import eq

def f(L, x):
    groups = groupby(L, partial(eq, 0))
    return list(chain.from_iterable(islice(v, x) if k else v for k,v in groups))