Two Sum function not working with recurring elements in list - python-3.x

Question

I'm trying to complete "Two Sum", which goes as such:
Write a function that takes an array of numbers (integers for the tests) and a target number. It should find two different items in the array that, when added together, give the target value. The indices of these items should then be returned in a tuple like so: (index1, index2).
Efficiency of my code aside, this is what I have so far:
def two_sum(numbers, target):
for i in numbers:
for t in numbers:
if i + t == target:
if numbers.index(i) != numbers.index(t):
return (numbers.index(i), numbers.index(t))
return False
It works for inputs such as:
>>> two_sum([1,2,3,4,5,6,7,8,9,10], 11)
(0, 9)
But when I try a list of numbers that have recurring numbers that add up to the target, the code doesn't work:
>>> two_sum([2, 2], 4)
False
The code, for some reason that I cannot figure out, does not reach index [1] of the list, and thus returns False.
Why is that?

Answer 1

The list method index() always returns the first occurence of an item in a list, so numbers.index(i) != numbers.index(t) evaluates to 1 != 1 which is False.
You should use the builtin enumerate() to store the indices while looping over the list.
def two_sum(numbers, target):
for i, number_a in enumerate(numbers):
for j, number_b in enumerate(numbers):
if number_a + number_b == target and i != j:
return (i, j)
return False

Answer 2

'''
return will break the loop and come out of function, so first you need to complete the cycle, store the result in list as you cant write to tuple,
once your loop gets completed convert list to tuple and return
'''
def two_sum(numbers, target):
result = []
for i in numbers:
for t in numbers:
if (i + t == target) and (numbers.index(i) != numbers.index(t)):
result.append(i)
result.append(t)
if (len(result)> 0):
return tuple(result)
else:
return False

Answer 3

Your code looks fine except this part:
if numbers.index(i) != numbers.index(t):
return (numbers.index(i), numbers.index(t))
return False
Because the index method returns only the first occurrence of a value, i and t are always the same. It will always return false. The index of the value 2 is always 0 in the list even though there is another 2 at index 1.
Source: https://www.w3schools.com/python/ref_list_index.asp
What you want to do is this:
def two_sum(numbers, target):
i_index = 0
t_index = 0
for i in numbers:
for t in numbers:
if i + t == target:
if i_index != t_index:
return (i_index, t_index)
t_index +=1
i_index +=1
return False
This way the index is not associated with the value

Answer 4

def pairs_sum_to_target(list1, list2, target):
'''
This function is about a game: it accepts a target integer named target and
two lists of integers (list1 and list2).
Then this function should return all pairs of indices in the form [i,j]
where list1[i] + list[j] == target.
To summarize, the function returns the pairs of indices where the sum of
their values equals to target.
Important: in this game list1 and list2 will always have the same number of
elements and returns the pairs in that order.
'''
pairs = [] #make a list, which is empty in the beginning. But store the sum pairs == target value.
#loop for all indices in list1 while looping all the same indices in list2 and comparing if the sum == target variable.
for i, value1 in enumerate (list1):
for j, value2 in enumerate(list2):
if value1 + value2 == target: ## if the value of element at indice i + value of element at indice j == target, then append the pairs to list pairs []- in order.
pairs.append(i,j)
return pairs
Simple Input #1
"""This is one example of input for list1, list2, and target. In order to properly test this function"""
list1 = [1,-2,4,5,9]
list2 = [4,2,-4,-4,0]