traverse python int lists by skipping consecutive numbers

Question

Guys, I’m trying to eliminate consecutive numbers in a list of integers. When the list is for example 1,2,3,5,6,7,8,9... it show: 1-3, 5-9 I tried many ways and I arrived at that code:

def apresenta(lst):

for i in lst:
    x = i #1
    a = i+1 #2 
    b = i+2 #3
    c = i+3 #4
    d = i+4 #5
    e = i+5 #6
    f = i+6 #7
    g = i+7 #8
    h = i+8 #9
    z = i+9 #10
    j = i+10 #11
    l = i+11 #12
    m = i+12 #13
    if len(lst) > 2:
        while b > x:
            lst.pop(a)
        while d > b:
            lst.pop(a)
        while f > d:
            lst.pop(e)
        while h > f:
            lst.pop(g)
        while j > h:
            lst.pop(z)
    return lst

when tested on presents([1,2,3,4])

He tells me Indexerror: pop index out of range

Someone can give me a light?

I know this code lacks much logic, but after several attempts I’m lost...

Answer 1

If I understand your doubt, follow the script to do what you need:

from itertools import groupby, count

l = [1,2,3,4,5,6,8,10,11]

l = list(set(l)) # Remove números duplicados
l.sort() # Ordenar os números

def apresenta(iterable):
  return ','.join(as_range(g) for _, g in groupby(l, key=lambda n, c=count(): n-next(c)))

def as_range(iterable):
    l = list(iterable)
    if len(l) > 1:
        return '{0}-{1}'.format(l[0], l[-1])
    else:
        return '{0}'.format(l[0])


print(apresenta(l))

The two commented lines can be used or not, depending on your need.