How do I loop through a list by twos? [duplicate]

I want to loop through a Python list and process 2 list items at a time. Something like this in another language:

for(int i = 0; i < list.length(); i+=2)
{
   // do something with list[i] and list[i + 1]
}

What's the best way to accomplish this?

You can use for in range with a step size of 2:

Python 2

for i in xrange(0,10,2):
  print(i)

Python 3

for i in range(0,10,2):
  print(i)

Note: Use xrange in Python 2 instead of range because it is more efficient as it generates an iterable object, and not the whole list.

You can also use this syntax (L[start:stop:step]):

mylist = [1,2,3,4,5,6,7,8,9,10]
for i in mylist[::2]:
    print i,
# prints 1 3 5 7 9

for i in mylist[1::2]:
    print i,
# prints 2 4 6 8 10

Where the first digit is the starting index (defaults to beginning of list or 0), 2nd is ending slice index (defaults to end of list), and the third digit is the offset or step.

The simplest in my opinion is just this:

it = iter([1,2,3,4,5,6])
for x, y in zip(it, it):
    print x, y

Out: 1 2
     3 4
     5 6

No extra imports or anything. And very elegant, in my opinion.

If you're using Python 2.6 or newer you can use the grouper recipe from the itertools module:

from itertools import izip_longest

def grouper(n, iterable, fillvalue=None):
    "grouper(3, 'ABCDEFG', 'x') --> ABC DEF Gxx"
    args = [iter(iterable)] * n
    return izip_longest(fillvalue=fillvalue, *args)

Call like this:

for item1, item2 in grouper(2, l):
    # Do something with item1 and item2

Note that in Python 3.x you should use zip_longest instead of izip_longest.

nums = range(10)
for i in range(0, len(nums)-1, 2):
    print nums[i]

Kinda dirty but it works.