How do I loop through a list by twos?

Question

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?

Answer 1

You can use a 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.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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

Kinda dirty but it works.

Answer 6

This might not be as fast as the izip_longest solution (I didn't actually test it), but it will work with python < 2.6 (izip_longest was added in 2.6):

from itertools import imap

def grouper(n, iterable):
    "grouper(3, 'ABCDEFG') --> ('A,'B','C'), ('D','E','F'), ('G',None,None)"
    args = [iter(iterable)] * n

    return imap(None, *args)

If you need to go earlier than 2.3, you can substitute the built-in map for imap. The disadvantage is that it provides no ability to customize the fill value.

Answer 7

If you have control over the structure of the list, the most pythonic thing to do would probably be to change it from:

l=[1,2,3,4]

to:

l=[(1,2),(3,4)]

Then, your loop would be:

for i,j in l:
    print i, j