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Newsgroups: comp.lang.python
Date: Sun, 30 Sep 2012 17:58:39 -0700 (PDT)
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Subject: Re: Slicing iterables in sub-generators without loosing elements
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Xref: csiph.com comp.lang.python:30579

On Sunday, September 30, 2012 12:15:57 AM UTC+8, Thomas Bach wrote:
> Hi,
> 
> 
> 
> say we have the following:
> 
> 
> 
> >>> data = [('foo', 1), ('foo', 2), ('bar', 3), ('bar', 2)]
> 
> 
> 
> is there a way to code a function iter_in_blocks such that
> 
> 
> 
> >>> result = [ list(block) for block in iter_in_blocks(data) ]
> 
> 
> 
> evaluates to
> 
> 
> 
> >>> result = [ [('foo', 1), ('foo', 2)], [('bar', 3), ('bar', 2)] ]
> 
> 
> 
> by _only_ _iterating_ over the list (caching all the elements sharing
> 
> the same first element doesn't count)?
> 
> 
> 
> I came up with the following
> 
> 
> 
> def iter_in_blocks(iterable):
> 
>     my_iter = iter(iterable)
> 
>     while True:
> 
>         first = next(my_iter)
> 
>         pred = lambda entry: entry[0] == first[0]
> 
>         def block_iter():
> 
>             yield first
> 
>             for entry in itertools.takewhile(pred, my_iter):
> 
>                 yield entry
> 
>         yield block_iter()
> 
> 
> 
> which does not work as itertools.takewhile consumes the first entry
> 
> not fulfilling the pred.
> 
> 
> 
> I currently have the intuition that the problem is not solvable
> 
> without using e.g. a global to pass something back to iter_in_blocks
> 
> from block_iter. Any other suggestions?
> 
> 
> 
> Regards,
> 
> 	Thomas Bach.

Your question seems vague to me. If you know you are storing 
only immutable tuples in a list, then the way to iterate is simple. 

For example: 

data = [('foo', 1), ('foo', 2), ('bar', 3), ('bar', 2)] 
# all tuples 

for item in data: 
    x1=item[0] # first entry in each tuple
    x2=item[1]
    print x1, x2 # or use yield in a function to iterate