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. -- http://mail.python.org/mailman/listinfo/python-list
