return a list of numbers that are in sequence

Question

I’m trying to make some list go through and make another one come back with the numbers that are in sequence.

My list is returning empty.

Ex: lista=[1,2,3,5,9,10] returns nova_lista=[1,2,3,9,10]

sequencial= []
def seq (lista):
     for i in range(0,len(lista)-1):
         if lista[i] == lista[i+1]:
             sequencial.append[i]
     return sequencial

Answer 1

The problem is that if lista[i] == lista[i+1] compares whether the element is equal to the next. And since the list has no repeated elements, the result will always be empty (as it will never enter this if).

A solution would be to scroll through the list from the second element and check if it is equal to "previous + 1":

lista = [1, 2, 3, 5, 9, 10]
result = []
in_seq = False
for i in range(1, len(lista)):
    if lista[i] == lista[i - 1] + 1:
        if not in_seq:
            result.append(lista[i - 1])
            in_seq = True
        result.append(lista[i])
    else:
        in_seq = False

print(result)

I also used the variable in_seq to know if I’m in the middle of a sequence, because if I’m not, I also have to include the previous element in the result.

The result is:

[1, 2, 3, 9, 10]