how convert list of int to list of tuples

Question

I want to convert a list like this

l1 = [1,2,3,4,5,6,7,8]

to

l2 = [(1,2),(3,4),(5,6),(7,8)]

because want to loop

for x,y in l2:
    draw_thing(x,y)

Answer 1

What's wrong with just accessing the correct index and incrementing?

for (int i=0;i<myList.Length;i++)
{
  draw_thing(myList[i],myList[++i]);
}

Oops - sorry, in C# mode. I'm sure you get the idea.

Answer 2

You can do:

l2 = []
for y in range(0, len(l1), 2):
    l2.append((l1[y], l1[y+1]))

I'm not doing any checks to make sure l1 has an even number of entries and such-like.

Answer 3

Not the most elegant solution

l2 = [(l1[i], l1[i+1]) for i in xrange(0,len(l1),2)]

Answer 4

One good way is:

from itertools import izip
it = iter([1, 2, 3, 4])
for x, y in izip(it, it):
    print x, y

Output:

1 2
3 4
>>>

Answer 5

No need to construct a new list. You can just iterate over the list by steps of 2 instead of 1. I use len(L) - 1 as the upper-bound so you ensure that you don't try to access past the end of the list.

for i in range(0, len(L) - 1, 2):
    draw_thing(L[i], L[i + 1])

Answer 6

Building on Nick D's answer:

>>> from itertools import izip
>>> t = [1,2,3,4,5,6,7,8,9,10,11,12]
>>> for a, b in izip(*[iter(t)]*2):
...     print a, b
...
1 2
3 4
5 6
7 8
9 10
11 12
>>> for a, b, c in izip(*[iter(t)]*3):
...     print a, b, c
...
1 2 3
4 5 6
7 8 9
10 11 12
>>> for a, b, c, d in izip(*[iter(t)]*4):
...     print a, b, c, d
...
1 2 3 4
5 6 7 8
9 10 11 12
>>> for a, b, c, d, e, f in izip(*[iter(t)]*6):
...     print a, b, c, d, e, f
...
1 2 3 4 5 6
7 8 9 10 11 12
>>>

Not quite as readable, but it shows a compact way to get any size tuple you want.

Answer 7

Take a look at grouper function from itertools docs.

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)

In your case use it like this:

l1 = [1,2,3,4,5,6,7,8]
for (x, y) in grouper(2, l1):
    draw_thing(x, y)

Answer 8

Kind of easy with python's slicing operator:

l2 = zip(l1[0::2], l1[1::2])

Answer 9

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