Iterability in Python

Question

I am trying to understand Iterability in Python.

As I understand, __iter__() should return an object that has next() method defined which must return a value or raise StopIteration exception. Thus I wrote this class which satisfies both these conditions.

But it doesn't seem to work. What is wrong?

class Iterator:
    def __init__(self):
        self.i = 1

    def __iter__(self):
        return self

    def next(self):
        if self.i < 5:
            return self.i
        else:
            raise StopIteration

if __name__ == __main__:
    ai = Iterator()
    b  = [i for i in ai]
    print b

Your current version has a space inside 'main', which is incorrect. — Van Gale
– Van Gale, Commented Mar 16, 2009 at 7:47

Haes · Accepted Answer · 2009-03-16 07:38:39Z

9

Your Iterator class is correct. You just have a typo in this statement:

if __name__ ==' __main__':

There's a leading whitespace in the ' __main__' string. That's why your code is not executed at all.

answered Mar 16, 2009 at 7:38

Haes

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Comments

cobbal · Accepted Answer · 2009-03-16 07:34:12Z

4

i will never become greater than 5 if you don't increment it in next()

edited Mar 16, 2009 at 7:34

answered Mar 16, 2009 at 7:27

cobbal

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Comments

Michael Kuhn · Accepted Answer · 2009-03-16 10:28:03Z

4

I think in most cases it might be enough to write a generator function that uses yield instead of writing a full-fledged iterator.

answered Mar 16, 2009 at 10:28

Michael Kuhn

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Comments

SingleNegationElimination · Accepted Answer · 2009-03-16 14:36:31Z

1

Your current code seems to work. Instead i'll show you some more iterators/generators.

the simplest builtin with exactly your behavior.

Iterator2 = xrange(2,5)

A direct translation of your class to a generator

def Iterator3():
    i = 1
    while i < 5:
        i += 1
        yield i

a generator composed from generators in the python standard library

import itertools
Iterator4 = itertools.takewhile( 
                        lambda y : y < 5, 
                        itertools.count(2) )

a simple generator expression (not very exciting...)

Iterator5 = ( x for x in [2, 3, 4] )

answered Mar 16, 2009 at 14:36

SingleNegationElimination

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Comments

Ethan Furman · Accepted Answer · 2016-03-18 23:19:08Z

0

Your code has two problems:

if name == '__main__': (missing quotes)
def next . . .: you don't increment i anywhere

answered Mar 18, 2016 at 23:19

Ethan Furman

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