about basic Python list indexing

Question

I'm not sure if this has been asked before as it seems to be a simple question, I tried to search it but did not find a very good answer.

Say I have a list of numbers a. I know I can get out every fifth number by a[::5].

What should be the simplest way to get the rest of the list (items in a but not in a[::5])?

Edit:
this is not the same question as Finding elements not in a list as I already know that the rest of the list consists of a[1::5]+a[2::5]+a[3::5]+a[4::5], I'm just asking is there a simpler way / syntax sugar for that?

Edit:
Thank you all for answering.
Is there a solution without scanning over the whole list?

Answer 1

I would start by something like:

[v for v in a if v not in a[::5]]

Store the unwanted value in a set before could lead to better perf, because not in method is in O(1) for sets:

unwanted = set(a[::5])
[v for v in a if v not in unwanted]

If the amount of data is huge, bloom filter could be a good substitute.

Answer 2

[x for i, x in enumerate(a) if i % 5 != 0]

not in set(a[::5]) will not work correctly when list contains equal elements. not in a[::5] is also not optimal from performance point of view.

Edit There is no syntactic sugar for your problem in Python. Take a look at numpy: it has extensive slicing possibilities.

Edit To avoid list scanning, you could create wrapper class with __getitem__ method.

Answer 3

Does this help you?

items = [1,2,3,4]
Z = [3,4,5,6]

a = list(set(items)-set(Z))
# a =[1, 2]

Such as:

fifths = a[::5]
not_fifths = list(set(a)-set(fifths))

Answer 4

temp = [x for x in a if x not in a[::5]]

Answer 5

You can try this:

[a[i-1] for i in range(1, len(a) + 1) if i % 5]

Basically, it checks for every position to see if it is a multiple of 5 or not. You don't have to create a set object this way.

Answer 6

This would do it, without any index look up overheads.

result = [n for i,n in enumerate(a) if i % 5 != 0]