[Baypiggies] a python puzzle

[Jason Culverhouse]

If the lists are already sorted wouldn't

	sorted(itertools.chain(a,b))

be fairly efficient?
http://bugs.python.org/file4451/timsort.txt

Jason


On Apr 14, 2010, at 9:03 AM, Emile van Sebille wrote:

> a = [1, 1, 2, 4, 6, 8, 9]
> b = [1, 3, 4, 5, 6, 7, 8]
> 
> def popwhile(l,t):
>  r = []
>  while l[:1] == t:
>    r.append(l.pop(0))
>  return r
> 
> while bool(a or b):
>  m = min(a[:1],b[:1]) or max(a[:1],b[:1])
>  print popwhile(a,m), popwhile(b,m)
> 
> 
> Emile
> 
> On 4/14/2010 8:20 AM Brent Pedersen said...
>> hi, in trying to write a func that does a kind of merging of 2 sorted lists,
>> i've come up with a fairly simple implementation that "almost works":
>> http://gist.github.com/365485
>> 
>> but it hits StopIteration before returning the last value ([9], None)
>> i can wrap the whole thing in a bunch more if statements, i've tried
>> heapq.merge, but cant find a nice solution.
>> 
>> any ideas? i think it's an interesting problem.
>> -brent
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