The word counter problem but instead using in-built iterator's __next__() function. So happy

trial.py

class counter():
    def get_next(self):

        while self.controller==True and self.i!=None:
            try:
                self.i=self.numbers.__next__()
            except:
                return None

            self.controller=True

            self.c=""
            while not self.i.isspace() and self.i!="\n" and self.i!=None :

                self.c += self.i
                self.controller=False
                try:
                    self.i = self.numbers.__next__()
                except:
                    return None

        self.controller=True
        print(self.c)
        return self.c

    def __init__(self,numbers):
        self.j=0
        self.i=""
        self.c=""
        self.controller=True
        self.store=[]
        self.numbers=iter(numbers)
        self.word= self.get_next()




    def __next__(self):
        n=0
        if self.word is None:
            raise StopIteration
        final=self.word
        # if word is None
        # print(word)

        self.store.append(self.word)
        for item in self.store:
            print(item)
            if self.word == item and self.word != None:
                n+=1
            elif self.word == None:
                return (final)
        self.word = self.get_next()
        print((final, n))
        print(self.store)
        return ((final,n))
    def __iter__(self):
        return self
numbers = """  
    one  
    one two  
    one two three   
    one two three four  
    one two three four five 
    one two three four five six
   """
mywords= counter(numbers)
r=[]

for item in mywords:
    r.append(item)
print(r)
print (mywords.__next__())