How to first remove every second element on a list, then every third on whats remaining?

Question

Say:

list = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

I know that list[::2] would remove every second element, so list = [1,3,5,7,9] What if say I then needed to remove every third element? So list would become [1,3,7,9] (5 would be removed since it is the third element. How would I then proceed to do that? Currently, using b = list[::3] returns [1, 7]

Answer 1

To delete elements from a given list, use del:

del lst[::2]  # delete every second element (counting from the first)
del lst[::3]  # delete every third

Demo:

>>> lst = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
>>> del lst[::2]
>>> lst
[2, 4, 6, 8, 10]
>>> del lst[::3]
>>> lst
[4, 6, 10]

If you wanted to delete the second element counting from the second, you'd need to give the slice a starting index other than the default:

del lst[1::2]  # delete every second element, starting from the second
del lst[2::3]  # delete every third element, starting from the third

Demo:

>>> lst = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
>>> del lst[1::2]
>>> lst
[1, 3, 5, 7, 9]
>>> del lst[2::3]
>>> lst
[1, 3, 7, 9]

Answer 2

Your initial statement saves every other value it does not delete it. You should also not make your variable "list".

given the original values you show above

del mylist[::2]

will return the even values of the list while

del mylist[1::2]

will return the odd values as you request. Following that the standard

del mylist[::3] 

will remove the third value of the list as you wanted.

Answer 3

For the incredibly long one-liner:

>>> [el for i,el in enumerate([el for i,el in enumerate([1,2,3,4,5,6,7,8,9,10]) if (i+1)%2]) if (i+1)%3]
[1,3,7,9]

Pseudocode for the above:

for (index, value) in [1,2,3,4,5,6,7,8,9,10]:
    if index+1 is divisible by 2: toss it
    else: add it to new_list

for (index, value) in new_list:
    if index+1 is divisible by 3: toss it
    else: add it to final_list

print(final_list)