Is assigning two variables to items in the same list the best way to access and perform operations on those items?

Question

Say I have a list

l = [1,2,3]

and I want to sum each item in the list with every other item in the list. I could do this:

x = [(a, b) for a in l for b in l]
y = [(a + b) for a in l for b in l]

x = [(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)]
y = [2, 3, 4, 3, 4, 5, 4, 5, 6]

Is assigning two variables (a and b) to items in the same list (l) the best way to access and perform operations on those items? Is there a better way to do this?

I've looked at list methods and functions, and couldn't find any.

UPDATE

Many people recommended the itertools product function so I thought I'd include timings before I mark as answered:

mysetup = """
from itertools import product
l = [10, 15, 3, 7]
k = 17
"""
mycode = """
[i for i in product(l, repeat=2)]
"""

t = timeit.timeit(setup=mysetup, stmt=mycode, number=100000)
# 0.1505305


mysetup2 = """
l = [10, 15, 3, 7]
k = 17
"""
mycode2 = """
[(a, b) for a in l for b in l]
"""

t1 = timeit.timeit(setup=mysetup2, stmt=mycode2, number=100000)
# 0.1432976

Answer 1

Python provides built in method

from itertools import product
l = [1,2,3]

Then generate the sum using list comprehension in a single step to be more efficent

result= [sum(i) for i in product(l, repeat= 2) ]

#result=[2, 3, 4, 3, 4, 5, 4, 5, 6]

Answer 2

Here's an itertools based approach:

from operator import add
from itertools import product, starmap
l = [1,2,3]

list(starmap(add, product(l, repeat=2)))
# [2, 3, 4, 3, 4, 5, 4, 5, 6]

---

Let's check the timings:

l = list(range(1000))

%timeit list(map(sum, product(l, repeat=2)))
# 187 ms ± 14.4 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%timeit [sum(combination) for combination in combinations_with_replacement(l, 2)]
# 123 ms ± 4.32 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

list(starmap(add, product(l, repeat=2)))
# 102 ms ± 4.11 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

Answer 3

Try this, itertools.product

>>> from itertools import product
>>> l = [1,2,3]
>>> [sum(combination) for combination in list(product(l, repeat = 2))]
[2, 3, 4, 3, 4, 5, 4, 5, 6]