py> from __future__ import generators
Hello again,
I was thinking about the __iter__ and what you were saying about
generators, so I opened up pyshell and started typing.
py> class R3:
...def __init__(self, d):
... self.d=d
... self.i=len(d)
...def __iter__(self):
... d,i
Terry ,
Thank you for the explanation . That is much clearer now, I have played
a bit with generators but have never tried to create a custom iterator.
I am just now getting into the internals part of python objects... this
langauage is still amazing to me!
The reason I asked the question was becau
"M.E.Farmer" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
> Terry Reedy wrote:
>>will. Try ip=iter(p) followed by ip.next and ip.next() instead.
> Does that mean if you dont't call iter(() on your instance or have a
> next() method you can't do this:
> p=R3('eggs')
> for
On Fri, 31 Dec 2004 16:31:45 GMT, Steven Bethard
<[EMAIL PROTECTED]> wrote:
>py> p = R3('eggs')
>py> i = p.__iter__()
>py> i.next()
>'s'
>or
>py> p = R3('eggs')
>py> i = iter(p)
>py> i.next()
>'s'
And that is precisely what I needed to know. Thanks, to you,
Terry and everyone who took time to l
Terry Reedy wrote:
>This is the wrong test for what I and some others thought you were
asking.
>The requirement for p to be an *iterable* and useable in code such as
'for
>i in p' is that iter(p), not p itself, have a .next method, and
iter(p)
>will. Try ip=iter(p) followed by ip.next and ip.next()
Bulba! wrote:
Thanks to everyone for their responses, but it still doesn't work re
returning next() method:
class R3:
def __init__(self, d):
self.d=d
self.i=len(d)
def __iter__(self):
d,i = self.d, self.i
while i>0:
i-=1
yield d[i]
"Bulba!" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
> Thanks to everyone for their responses, but it still doesn't work re
> returning next() method:
> class R3:
>def __init__(self, d):
>self.d=d
>self.i=len(d)
>def __iter__(self):
>d,i = self
Reread Russel Blau post he is spot on with his comments:
Russel Blau wrote:
>I don't get that from the passage quoted, at all, although it is
somewhat
>opaque. It says that your __iter__() method must *return an object*
with a
>next() method; your __iter__() method below doesn't return such an
obje
On Thu, 30 Dec 2004 12:06:31 -0800, Scott David Daniels
<[EMAIL PROTECTED]> wrote:
>Here's one way: # (Make __iter__ an iterator)
>Py> class R1(object):
> def __init__(self, data):
> self.data = data
> self.i = len(data)
> def __iter__(self):
>
Bulba! wrote:
Hello Mr Everyone,
From:
http://docs.python.org/tut/node11.html#SECTION001190
"Define a __iter__() method which returns an object with a next()
method. If the class defines next(), then __iter__() can just return
self:"
The thing is, I tried to define __iter__() direct
"Bulba!" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
> "Define a __iter__() method which returns an object with a next()
> method. If the class defines next(), then __iter__() can just return
> self:"
>
> The thing is, I tried to define __iter__() directly without explicit
> defin
Bulba! <[EMAIL PROTECTED]> wrote:
> So which is it? Does next() method HAS to be defined
> explicitly?
It has to be defined, whether explicitly or not (e.g. via a generator).
Alex
--
http://mail.python.org/mailman/listinfo/python-list
"Bulba!" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
> Hello Mr Everyone,
>
> From:
> http://docs.python.org/tut/node11.html#SECTION001190
>
> "Define a __iter__() method which returns an object with a next()
> method. If the class defines next(), then __iter__() can
Hello Mr Everyone,
From:
http://docs.python.org/tut/node11.html#SECTION001190
"Define a __iter__() method which returns an object with a next()
method. If the class defines next(), then __iter__() can just return
self:"
The thing is, I tried to define __iter__() directly without
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