Return a non iterator from __iter__

Question

class test(object):
    def __init__(self):
        pass
    def __iter__(self):
        return \"my string\"

o = test()
print iter(o)

Why does th

Answer 1

str is an iterable but not an iterator, subtle but important difference. See this answer for an explanation.

You want to return an object with __next__ (or just next if py2) which is what str returns when is iterated.

def __iter__(self):
      return iter("my string")

str does not implement __next__

In [139]: s = 'mystring'

In [140]: next(s)
---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
 in ()
----> 1 next(s)

TypeError: 'str' object is not an iterator

However calling iter returns the iterator, which is what looping calls:

In [141]: next(iter(s))
Out[141]: 'm'

You get the same problem returning anything without a __next__ (or next in py2) method

You can use a generator, which itself has __iter__ that returns self:

def gen():
    yield 'foo'

gg = gen()

gg is gg.__iter__()
True

gg.__next__()
'foo'

class Something:
     def __iter__(self):
         return gen()

list(Something())
['foo']

Or a class where you implement __next__ yourself, like this class similar to the one on the Ops post (you also have to handle StopIteration which stops the loop)

class test:
    def __init__(self, somestring):
        self.s = iter(somestring)

    def __iter__(self):
        return self

    def __next__(self):
        return next(self.s) ## this exhausts the generator and raises StopIteration when done.

In [3]: s = test('foo')

In [4]: for i in s:
   ...:     print(i)
   ...:
f
o
o