Python iterator returning unwanted 'None'

Question

Why is my iterator returning extra \'None\' in the output. For the parameters/example below, I am getting [None,4,None] instead of the desired [4] Ca

Answer 1

Functions return None if you don't have an explicit return statement. That's what happens in __next__ when if (self.purchase[old])%(self.d) == 0: isn't true. You want to stay in your __next__ until it has a value to return.

class Prizes(object):
    def __init__(self,purchase,n,d):
        self.purchase = purchase
        self.length = len(purchase)
        self.i = n-1
        self.n = n
        self.d = d

    def __iter__(self):
        return self

    def __next__(self):
        while self.i < self.length:
            old = self.i
            self.i += self.n
            if (self.purchase[old])%(self.d) == 0:
                return old+1
        raise StopIteration

def superPrize(purchases, n, d):
    return list(Prizes(purchases, n, d))

purchases = [12, 43, 13, 465, 1, 13]
n = 2
d = 3
print(superPrize(purchases, n, d))

Answer 2

As people in the comments have pointed out, your line if (self.purchase[old])%(self.d) == 0: leads to the function returning without any return value. If there is no return value supplied None is implied. You need some way of continuing through your list to the next available value that passes this test before returning or raising StopIteration. One easy way of doing this is simply to add an extra else clause to call self.__next__() again if the test fails.

def __next__(self):
        if self.i < self.length:
            old = self.i
            self.i += self.n
            if (self.purchase[old])%(self.d) == 0:
                print("returning")
                return old+1
            else:
                return self.__next__()
        else:
            raise StopIteration