iterator question

[Steve Holden]

George Sakkis wrote:
> Neil Cerutti wrote:
> 
> 
>>On 2006-09-26, Neal Becker <ndbecker2 at gmail.com> wrote:
>>
>>>Any suggestions for transforming the sequence:
>>>
>>>[1, 2, 3, 4...]
>>>Where 1,2,3.. are it the ith item in an arbitrary sequence
>>>
>>>into a succession of tuples:
>>>
>>>[(1, 2), (3, 4)...]
>>>
>>>In other words, given a seq and an integer that specifies the
>>>size of tuple to return, then for example:
>>
>>It turns out there's a itertools recipe to do this; the last one
>>in the itertools recipe book:
>>
>>def grouper(n, iterable, padvalue=None):
>>    """
>>    grouper(3, 'abcdefg', 'x') --> ('a','b','c'), ('d','e','f'), ('g','x','x')
>>
>>    """
>>    return izip(*[chain(iterable, repeat(padvalue, n-1))]*n)
> 
> 
> That's not quite the same as the previous suggestions; if the last
> tuple is shorter than n, it pads the last tuple with padvalue. The OP
> didn't mention if he wants that or he'd rather have a shorter last
> tuple.
> 
In which case why not go in for a bit of requirements gold-plating and 
add a keyword Boolean argument that allows you to specify which 
behaviour you want. Or, alternatively, let the OP make sense of the 
suggestions already made.

regards
  Steve
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