how to make a generator use the last yielded value when it regains control

[John Salerno]

Gerard Flanagan wrote:
> John Salerno wrote:
> 
>> Michael Spencer wrote:
>>
>>> itertools.groupby makes this very straightforward:
>> I was considering this function, but then it seemed like it was only
>> used for determing consecutive numbers like 1, 2, 3 -- not consecutive
>> equivalent numbers like 1, 1, 1. But is that not right?
> 
> 
> data = [1, 1, 1, 2, 2, 3, 4, 4, 3, 2, 2, 1, 1, 2, 2,4, 2, 2]
> 
> from itertools import groupby
> 
> 
> for k, g in groupby( data ):
>     print k, list(g)
> 
> 1 [1, 1, 1]
> 2 [2, 2]
> 3 [3]
> 4 [4, 4]
> 3 [3]
> 2 [2, 2]
> 1 [1, 1]
> 2 [2, 2]
> 4 [4]
> 2 [2, 2]
> 
> for k, g in groupby( data, lambda x: x<2 ):
>     print k, list(g)
> 
> True [1, 1, 1]
> False [2, 2, 3, 4, 4, 3, 2, 2]
> True [1, 1]
> False [2, 2, 4, 2, 2]    
> 
> Gerard
> 

Interesting. After following along with the doc example, it seemed like 
   I had to do complicated stuff with the keys parameter, and I kind of 
lost track of it all.