How to stop look returning None when there is an element that satisfies if condition?

Question

I am looping through a list of booleans my conditions are if the position of the element is > m and if the element = True

the function will return the position of the element

this is what I've done :

panda =[True, True, True, True]

def find_next (l, m): 
    for i in l:  
        if ((l.index(i) > m) and i ==True):
            return l.index(i) 

print(find_next(panda, 2))

I expected the output to be 3.

but I got None . why?

Answer 1

l.index(i) always returns 0 because it's finding the first instance of True in your list

As an aside, you don't need to call l.index on every loop because you are redundantly searching the list when you should already know what iteration you are on.

def find_next(l, m):
    for index, value in enumerate(l):
        if index > m and value:
            return index