Given a long number, return all the possible sum of two digits of it.

For example, `12345`: all possible sum of two digits from that number are:

    [ 1 + 2, 1 + 3, 1 + 4, 1 + 5, 2 + 3, 2 + 4, 2 + 5, 3 + 4, 3 + 5, 4 + 5 ]

Therefore the result must be:

    [ 3, 4, 5, 6, 5, 6, 7, 7, 8, 9 ]

Source: codewars | Difficulty: EASY

Starter code: def digits(num):
	

Solutions:
from itertools import combinations

def digits(num):
	return list(map(sum, combinations(map(int, str(num)), 2)))


def digits(num):
	return [int(a) + int(b) for (i, a) in enumerate(str(num)) for b in str(num)[i + 1:]]

def digits(num):
	from itertools import combinations
	return [int(a) + int(b) for (a, b) in combinations(str(num), 2)]


from itertools import combinations

def digits(num):
	return [int(a) + int(b) for (a, b) in combinations(str(num), 2)]


from itertools import combinations

def digits(num):
	return [sum(pair) for pair in combinations([int(d) for d in str(num)], 2)]


from itertools import combinations

def digits(num):
	ns = list(map(int, str(num)))
	return [a + b for (a, b) in combinations(ns, 2)]


def digits(num):
	num = str(num)
	return [int(x) + int(y) for (n, x) in enumerate(num, 1) for y in num[n:]]


def digits(num):
	return [int(x) + int(y) for (i, x) in enumerate(str(num)) for (j, y) in enumerate(str(num)) if j > i]


def digits(num):
	ints = [int(i) for i in str(num)]
	sums = []
	for i in range(len(ints) - 1):
		for j in ints[i + 1:]:
			sums.append(ints[i] + j)
	return sums


def digits(num):
	parse = lambda s: [int(s[0]) + int(x) for x in s[1:]] + (digits(s[1:]) if len(s) > 2 else [])
	return parse(str(num))