Taking list's tail in a Pythonic way?

Question

from random import randrange
data = [(randrange(8), randrange(8)) for x in range(8)]

And we have to test if the first item equals to one of a tail. I am curious, how we would do it in most simple way without copying tail items to the new list? Please take into account this piece of code gets executed many times in, say, update() method, and therefore it has to be quick as possible.

Using an additional list (unnesessary memory wasting, i guess):

head = data[0]
result = head in data[1:]

Okay, here's another way (too lengthy):

i = 1
while i < len(data):
    result = head == data[i]
    if result:
        break
    i+=1

What is the most Pythonic way to solve this? Thanks.

Answer 1

Nick D's Answer is better

use islice. It doesn't make a copy of the list and essentially embeds your second (elegant but verbose) solution in a C module.

import itertools

head = data[0]
result = head in itertools.islice(data, 1, None)

for a demo:

>>> a = [1, 2, 3, 1]
>>> head = a[0]
>>> tail = itertools.islice(a, 1, None)
>>> head in tail
True

Note that you can only traverse it once but if all you want to do is check that the head is or is not in the tail and you're worried about memory, then I think that this is the best bet.