Python “Every Other Element” Idiom [duplicate]

Answer 1

The one often-quoted is:

zip(*[iter(l)] * 2)

---

I prefer this more readable version of the iter solution:

it = iter(l)
list(zip(it, it))
# [(1, 2), (3, 4), (5, 6)]

Answer 2

The right thing is probably not to compute lists, but to write an iterator->iterator function. This is more generic -- it works on every iterable, and if you want to "freeze" it into a list, you can use the "list()" function.

def groupElements(iterable, n):
    # For your case, you can hardcode n=2, but I wanted the general case here.
    # Also, you do not specify what to do if the 
    # length of the list is not divisible by 2
    # I chose here to drop such elements
    source = iter(iterable)
    while True:
        l = []
        for i in range(n):
            l.append(source.next())
        yield tuple(l)

I'm surprised the itertools module does not already have a function for that -- perhaps a future revision. Until then, feel free to use the version above :)