Fix Python – If range() is a generator in Python 3.3, why can I not call next() on a range?


Asked By – Jeff

Perhaps I’ve fallen victim to misinformation on the web, but I think it’s more likely just that I’ve misunderstood something. Based on what I’ve learned so far, range() is a generator, and generators can be used as iterators. However, this code:

myrange = range(10)

gives me this error:

TypeError: 'range' object is not an iterator

What am I missing here? I was expecting this to print 0, and to advance to the next value in myrange. I’m new to Python, so please accept my apologies for the rather basic question, but I couldn’t find a good explanation anywhere else.

Now we will see solution for issue: If range() is a generator in Python 3.3, why can I not call next() on a range?


range is a class of immutable iterable objects. Their iteration behavior can be compared to lists: you can’t call next directly on them; you have to get an iterator by using iter.

So no, range is not a generator.

You may be thinking, “why didn’t they make it an iterator”? Well, ranges have some useful properties that wouldn’t be possible that way:

  • They are immutable, so they can be used as dictionary keys.
  • They have the start, stop and step attributes (since Python 3.3), count and index methods and they support in, len and __getitem__ operations.
  • You can iterate over the same range multiple times.

>>> myrange = range(1, 21, 2)
>>> myrange.start
>>> myrange.step
>>> myrange.index(17)
>>> myrange.index(18)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: 18 is not in range
>>> it = iter(myrange)
>>> it
<range_iterator object at 0x7f504a9be960>
>>> next(it)
>>> next(it)
>>> next(it)

This question is answered By – Oleh Prypin

This answer is collected from stackoverflow and reviewed by FixPython community admins, is licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0