Big Oh Time Complexity Involving 'for i in range(n)' Given the code below and the comment analysis:

n = len(L)  # O(1)
for i in range(n):  # O(n) - I read online that range(n) will take constant time 
                      but the 'for i in range(n)' part would make this O(n)?
    if L[i] &gt; 0 :  # O(1)
       for j in range(n):  # O(n)
           answer = answer + L[i]  # O(1) - since indexing takes constant time
    else :  # O(1)
        for j in range(n):  # O(n)
            answer = answer - L[i]   # O(1)


My Overall Analysis:


$O(1)$ for the first variable assignment
first if statement takes overall $O(n)$
else statement also takes overall $O(n)$
=> Overall if/else statements take $O(n)$ as well
for loop iterates n times => $O(n * n)$ => $O(n^2)$


=> Overall time complexity $O(n^2)$

Thanks in advance for the suggestions!

EDIT: I am wondering if the

for i in range(n) 


portion of the code takes O(n) or O(1)? I read that range(n) would take O(1) but since its a for loop it would take O(n)?
So you have the right logic, if you have a loop of $O(n)$ iterations, the complexity of the entire loop will be $O(n*\text{complexity of loop operations})$. In this case, you again are correct that your loop's complexity is $O(n)$ for each iteration. 

Your last bullet point shows that you understand this as well, as the total loop complexity is then $O(n^2)$. However, this trivially results in an overall time complexity of $O(n^2)$, not $O(n)$ as you concluded.

I imagine what you were really asking about is the complexity of range(n). While it is true that the initial call to range is $O(1)$, that is because it produces a generator which takes $O(1)$ to return the next successive value i.e. it doesn't produce the entire range right away, but instead on demand. However, keep in mind that the generator will be advanced $O(n)$ times (once for every iteration of the loop).