Help on for loop understanding

[Robert]

On Monday, December 7, 2015 at 8:32:30 PM UTC-5, Robert wrote:
> On Monday, December 7, 2015 at 8:14:46 PM UTC-5, Robin Koch wrote:
> > Am 08.12.2015 um 02:05 schrieb Robert:
> > > Hi,
> > > When I learn for loop with below link:
> > >
> > > http://www.shutupandship.com/2012/01/understanding-python-iterables-and.html
> > >
> > > it has such explanation:
> > >
> > > \\\\\\\\\
> > > for loop under the hood
> > >
> > > First let's look at the for loop under the hood. When Python executes the
> > >   for loop, it first invokes the __iter__() method of the container to get the
> > >   iterator of the container. It then repeatedly calls the next() method
> > >   (__next__() method in Python 3.x) of the iterator until the iterator raises a
> > >   StopIteration exception. Once the exception is raised, the for loop ends.
> > > \\\\\\\\\
> > >
> > > When I follow a list example from the above link, and one example of myself:
> > >
> > > //////////
> > > xx=[1,2,3,4,5]
> > >
> > > xx.next
> > > ---------------------------------------------------------------------------
> > > AttributeError                            Traceback (most recent call last)
> > > <ipython-input-76-dd0716c641b1> in <module>()
> > > ----> 1 xx.next
> > >
> > > AttributeError: 'list' object has no attribute 'next'
> > >
> > > xx.__iter__
> > > Out[77]: <method-wrapper '__iter__' of list object at 0x000000000A1ACE08>
> > >
> > > for c in xx: print c
> > > 1
> > > 2
> > > 3
> > > 4
> > > 5
> > > //////////////
> > >
> > > I am puzzled that the list examples have no next method, but it can run as
> > > a for loop. Could you explain it to me the difference?
> > 
> > Lists don't have a next method. Their iterators have:
> > 
> > xx.__iter__().__next__()
> > 
> > or
> > 
> > xxIterator = xx.__iter__()
> > xxIterator.__next__()
> > xxIterator.__next__()
> > xxIterator.__next__()
> > xxIterator.__next__()
> > xxIterator.__next__()
> > 
> > That's also what your quoted paragraph states:
> > 
> > | It then repeatedly calls the next() method
> > | (__next__() method in Python 3.x) of the iterator
> > 
> > -- 
> > Robin Koch
> 
> I use Python 2.7. I have tried these commands:
> 
> xx=[1,2,3,4,5]
> 
> xx.__iter__
> Out[2]: <method-wrapper '__iter__' of list object at 0x000000000A1B85C8>
> 
> xx.__iter__().__next__()
> ---------------------------------------------------------------------------
> AttributeError                            Traceback (most recent call last)
> <ipython-input-3-980bcf1bbf42> in <module>()
> ----> 1 xx.__iter__().__next__()
> 
> AttributeError: 'listiterator' object has no attribute '__next__' 
> 
> xx.__iter__.__next__
> ---------------------------------------------------------------------------
> AttributeError                            Traceback (most recent call last)
> <ipython-input-4-df8d8c6955e2> in <module>()
> ----> 1 xx.__iter__.__next__
> 
> AttributeError: 'method-wrapper' object has no attribute '__next__' 
> 
> xx.__iter__()
> Out[5]: <listiterator at 0xa211780>
> 
> xx.__iter__().__next__
> ---------------------------------------------------------------------------
> AttributeError                            Traceback (most recent call last)
> <ipython-input-6-92f21313d62e> in <module>()
> ----> 1 xx.__iter__().__next__
> 
> AttributeError: 'listiterator' object has no attribute '__next__' 
> 
> xxIterator = xx.__iter__()
> 
> xxIterator
> Out[8]: <listiterator at 0xa2115c0>
> 
> xxIterator.__next__()
> ---------------------------------------------------------------------------
> AttributeError                            Traceback (most recent call last)
> <ipython-input-9-5b74e35c2c6e> in <module>()
> ----> 1 xxIterator.__next__()
> 
> AttributeError: 'listiterator' object has no attribute '__next__' 
> 
> for c in xx: print c
> 1
> 2
> 3
> 4
> 5
> ------------
> 
> I don't find a way to show __next__ yet. 
> Can we explicitly get the iterator for a list?
> Thanks,

Excuse me. I find it as the following:

xx.__iter__().next
Out[16]: <method-wrapper 'next' of listiterator object at 0x0000000008B38AC8>

xx.__iter__().next()
Out[17]: 1