On Oct 21, 11:46 am, Yingjie Lan lany...@yahoo.com wrote:
I am still not sure why should we enforce that
a generator can not be reused after an explicit
request to revive it?
No one is enforcing anything, you're simply resisting implementing
this yourself. Consider the following generator:
On Oct 21, 12:09 pm, Yingjie Lan lany...@yahoo.com wrote:
Secondly, it would be nice to automatically revive it.
Sure, it's always nice when your expectation of a language feature
exactly matches with its capabilities.
When it doesn't, you suck it up and code around it.
Because at the very
in xrange(3))
else:
g = (x+x for x in xrange(3))
for y in g:
print x
revive(g) # which generator expression was it?
# need to carry around a reference to be able to tell
for y in g:
print x
Carrying a reference to a code object in turn carries around
Here's an example of an explicit request to revive the generator:
g = (x*x for x in range(3))
for x in g: print x
0
1
4
g = (x*x for x in range(3)) # revive the generator
for x in g: print x #now this will work
0
1
4
ChrisA
What if the generator is passed in as an argument
not be reused after an explicit
request to revive it?
The language has no explicit notion of a request to revive a
generator. You could use the same syntax to make a new generator that
yeilds the same values as the one you started with if that's what you
want.
As we've already discussed if you want
- Original Message -
From: Paul Rudin paul.nos...@rudin.co.uk
The language has no explicit notion of a request to revive a
generator. You could use the same syntax to make a new generator that
yeilds the same values as the one you started with if that's what you
want.
As we've
- Original Message -
From: Paul Rudin paul.nos...@rudin.co.uk
To: python-list@python.org
Cc:
Sent: Friday, October 21, 2011 3:27 PM
Subject: Re: revive a generator
The language has no explicit notion of a request to revive a
generator. You could use the same syntax to make
Yingjie Lan lany...@yahoo.com writes:
What if the generator involves a variable from another scope,
and before re-generating, the variable changed its value.
Also, the generator could be passed in as an argument,
so that we don't know its exact expression.
vo = 34
g = (vo*x for x in
On Fri, Oct 21, 2011 at 7:02 PM, Yingjie Lan lany...@yahoo.com wrote:
What if the generator involves a variable from another scope,
and before re-generating, the variable changed its value.
Also, the generator could be passed in as an argument,
so that we don't know its exact expression.
- Original Message -
From: Paul Rudin paul.nos...@rudin.co.uk
I'm not really sure whether you intend g to yield the original values
after your revive or new values based on the new value of vo. But
still you can make a class that supports the iterator protocol and does
whatever
- Original Message -
From: Chris Angelico ros...@gmail.com
To: python-list@python.org
Cc:
Sent: Friday, October 21, 2011 4:27 PM
Subject: Re: revive a generator
On Fri, Oct 21, 2011 at 7:02 PM, Yingjie Lan lany...@yahoo.com wrote:
What if the generator involves a variable
On 10/20/2011 10:09 PM, Yingjie Lan wrote:
snip
What if the generator is passed in as an argument
when you are writing a function? That is, the expression
is not available?
Secondly, it would be nice to automatically revive it.
For example, when another for-statement or other
equivalent is
On Thu, 20 Oct 2011 19:09:42 -0700, Yingjie Lan wrote:
Here's an example of an explicit request to revive the generator:
g = (x*x for x in range(3))
for x in g: print x
0
1
4
g = (x*x for x in range(3)) # revive the generator for x in g:
print x #now this will work
0
1
4
On Fri, Oct 21, 2011 at 2:02 AM, Yingjie Lan lany...@yahoo.com wrote:
Oops, my former reply has the code indentation messed up
by the mail system. Here is a reformatted one:
What if the generator involves a variable from another scope,
and before re-generating, the variable changed its
Here is a class that creates a re-iterable from any callable, such as a
generator function, that returns an iterator when called, + captured
arguments to be given to the function.
class reiterable():
def __init__(self, itercall, *args, **kwds):
self.f = itercall # callable that returns
On Fri, 21 Oct 2011 16:25:47 -0400, Terry Reedy wrote:
Here is a class that creates a re-iterable from any callable, such as a
generator function, that returns an iterator when called, + captured
arguments to be given to the function.
class reiterable():
def __init__(self, itercall,
Hi,
it seems a generator expression can be used only once:
g = (x*x for x in range(3))
for x in g: print x
0
1
4
for x in g: print x #nothing printed
Is there any way to revive g here?
Yingjie
--
http://mail.python.org/mailman/listinfo/python-list
On Fri, Oct 21, 2011 at 12:23 AM, Yingjie Lan lany...@yahoo.com wrote:
Hi,
it seems a generator expression can be used only once:
g = (x*x for x in range(3))
for x in g: print x
0
1
4
for x in g: print x #nothing printed
Is there any way to revive g here?
If you're not generating
Yingjie Lan lany...@yahoo.com writes:
Hi,
it seems a generator expression can be used only once:
g = (x*x for x in range(3))
for x in g: print x
0
1
4
for x in g: print x #nothing printed
Is there any way to revive g here?
Generators are like that - you consume them until they run
On 10/20/2011 9:23 AM, Yingjie Lan wrote:
it seems a generator expression can be used only once:
Generators are iterators. Once iterators raise StopIteration, they are
supposed to continue doing so.
A generator expression defines a temporary anonymous generator function
that is called once
- Original Message -
From: Paul Rudin paul.nos...@rudin.co.uk
To: python-list@python.org
Cc:
Sent: Thursday, October 20, 2011 10:28 PM
Subject: Re: revive a generator
Yingjie Lan lany...@yahoo.com writes:
Hi,
it seems a generator expression can be used only once:
g
the generator:
g = (x*x for x in range(3))
for x in g: print x
0
1
4
g = (x*x for x in range(3)) # revive the generator
for x in g: print x #now this will work
0
1
4
ChrisA
--
http://mail.python.org/mailman/listinfo/python-list
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