Circular list in Python

Just add the first element to the end of the remaining elements from index 4 to the end

l1  = li[1:4]
l2 = li[4:] + [li[0]]

In [22]: li = [1,2,3,4,5,6,7,8]

In [23]: l1  = li[1:4]

In [24]: l2 = li[4:]+ [li[0]]

In [25]: l1
Out[25]: [2, 3, 4]

In [26]: l2
Out[26]: [5, 6, 7, 8, 1]

In two steps using extend:

l2 = li[4:]    
In [30]: l2.extend([li[0]])  
Out[31]: [5, 6, 7, 8, 1]

The simplest (not the most efficient) way is to take the same list twice and slice from that:

def circ_slice(a, start, length):
    return (a * 2)[start:start+length]

a = [0, 1,2,3,4,5,6,7,8]
print circ_slice(a, 2, 4)  # [2, 3, 4, 5]
print circ_slice(a, 7, 4)  # [7, 8, 0, 1]

"tail + head" solutions require if statements if the start position is not known in advance.

Here's a more efficient version:

import itertools

def circ_slice(a, start, length):
    it = itertools.cycle(a)
    next(itertools.islice(it, start, start), None)
    return list(itertools.islice(it, length))

Try this:

l1  = li[1:4]
l2 = li[4:]+ [li[0]]

On a side note:

You may check itertools.cycle which can be used to get the result in an efficient way.

Make an iterator returning elements from the iterable and saving a copy of each. When the iterable is exhausted, return elements from the saved copy. Repeats indefinitely