All you need is list slicing starting from index 1 and then you need to jump by 3 indexes:

>>> my_list = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]
>>> my_list[1::3]
[2, 5, 8, 11]

General syntax of list slicing is your_list[start:end:jump] where:

• start: is the starting index for slicing the list. Empty value means start of the list i.e. index 0

• end: is the index till which you want to slice the list. Empty value means end of the list

• jump: is used to jump the elements from start such that you'll get values in the order start, start+jump, start+2*jump, so on till your list reaches end. Empty value means 1

As mentioned by Mr. cᴏʟᴅsᴘᴇᴇᴅ in comment, for large lists one should prefer itertools.islice which returns iterator object and could be used in this case as:

from itertools import islice  # import islice

islice(my_list, 1, None, 3)

Very rarely using slice objects is easier to read than employing a list comprehension, and this is not one of those cases.

>>> A = [[1,2,3,4,5],[1,2,3,4,5],[1,2,3,4,5]]
>>> [sublist[:3] for sublist in A]
[[1, 2, 3], [1, 2, 3], [1, 2, 3]]

This is very clear. For every sublist in A, give me the list of the first three elements.

that's what slice() is for:

a = range(10)
s = slice(2,4)
print a[s]

That's the same as using a[2:4].

The short answer

Slicing lists does not generate copies of the objects in the list; it just copies the references to them. That is the answer to the question as asked.

The long answer

Testing on mutable and immutable values

First, let's test the basic claim. We can show that even in the case of immutable objects like integers, only the reference is copied. Here are three different integer objects, each with the same value:

>>> a = [1000 + 1, 1000 + 1, 1000 + 1]

They have the same value, but you can see they are three distinct objects because they have different ids:

>>> map(id, a)
[140502922988976, 140502922988952, 140502922988928]

When you slice them, the references remain the same. No new objects have been created:

>>> b = a[1:3]
>>> map(id, b)
[140502922988952, 140502922988928]

Using different objects with the same value shows that the copy process doesn't bother with interning -- it just directly copies the references.

Testing with mutable values gives the same result:

>>> a = [{0: 'zero', 1: 'one'}, ['foo', 'bar']]
>>> map(id, a)
[4380777000, 4380712040]
>>> map(id, a[1:]
... )
[4380712040]

Examining remaining memory overhead

Of course the references themselves are copied. Each one costs 8 bytes on a 64-bit machine. And each list has its own memory overhead of 72 bytes:

>>> for i in range(len(a)):
...     x = a[:i]
...     print('len: {}'.format(len(x)))
...     print('size: {}'.format(sys.getsizeof(x)))
... 
len: 0
size: 72
len: 1
size: 80
len: 2
size: 88

As Joe Pinsonault reminds us, that overhead adds up. And integer objects themselves are not very large -- they are three times larger than references. So this saves you some memory in an absolute sense, but asymptotically, it might be nice to be able to have multiple lists that are "views" into the same memory.

Saving memory by using views

Unfortunately, Python provides no easy way to produce objects that are "views" into lists. Or perhaps I should say "fortunately"! It means you don't have to worry about where a slice comes from; changes to the original won't affect the slice. Overall, that makes reasoning about a program's behavior much easier.

If you really want to save memory by working with views, consider using numpy arrays. When you slice a numpy array, the memory is shared between the slice and the original:

>>> a = numpy.arange(3)
>>> a
array([0, 1, 2])
>>> b = a[1:3]
>>> b
array([1, 2])

What happens when we modify a and look again at b?

>>> a[2] = 1001
>>> b
array([   1, 1001])

But this means you have to be sure that when you modify one object, you aren't inadvertently modifying another. That's the trade-off when you use numpy: less work for the computer, and more work for the programmer!