Python iterator returning unwanted 'None' -


why iterator returning 'none' in output. parameters/example below, getting [none,4,none] instead of desired [4] can explain why getting none , how can fix it? print out 'returning' appears once assuming 1 item should appended returning calling function.

code:

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):         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:             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))

output:

returning [none, 4, none]

as people in comments have pointed out, line if (self.purchase[old])%(self.d) == 0: leads function returning without return value. if there no return value supplied none implied. need way of continuing through list next available value passes test before returning or raising stopiteration. 1 easy way of doing add else clause call self.__next__() again if 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

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