Help with dict and iter

[mattia]

Il Sun, 29 Mar 2009 11:17:50 -0400, andrew cooke ha scritto:

> mattia wrote:
>> Hi all, I a list of jobs and each job has to be processed in a
>> particular order by a list of machines.
>> A simple representation is:
>> # Ordering of machines
>> JOB1 = [3, 1, 2, 4]
>> JOB2 = [2, 3, 1, 4]
>> JOBS = [JOB1, JOB2]
>> NJOBS = len(JOBS)
>> Now, I have a list of jobs and I want to have the associated list of
>> machines, e.g:
>> [JOB1, JOB1, JOB2] --> [3, 1, 2]
>> My original idea was to have a dict with associated the job number and
>> an iterator associated by the list of machines: job_machine =
>> dict((x+1, iter(JOBS[x])) for x in range(NJOBS)) Now, something like:
>> for x in job_list:
>>     print(next(job_machine[x]))
>> Works good, but imagine I have a list of job_list, now obviously I have
>> a StopIteration exception after the
>> first list. So, I'm looking for a way to "reset" the next() value every
>> time i complete the scan of a list.
> 
> don't you just want to have a new job machine?
> 
> for job_list in job_list_list:
>   job_machine = dict((x+1, iter(JOBS[x])) for x in range(NJOBS)) for x
>   in job_list:
>     print(next(job_machine[x]))
> 
> you can certainly write a generator that resets itself - i've done it by
> adding a special reset exception and then using generator.throw(Reset())
> - but it's a lot more complicated and i don't think you need it.  if you
> do, you need to look at writing generators explicitly using yield, and
> then at "enhanced generators" for using throw().
> 
> andrew

Well, you are right, just creating every time a new dict is a good 
solution, I was just adding complexity to a simple problem, thanks.