Stop iterating over nested loops and continue with most outest loop

Question

I am coding a heuristic for an optimization problem from the field of production. In this heuristic I have various conditions, stop criteria etc. In order to account for these different criteria, I worked with multiple nested loops, as you can see in the code below:

for tao in PERIODS:
    print ("Iteration:", tao)
    print ("-----------------------------------------")
    print (SETUP_ITEMS)
    for z in range(1,periods_count+1-tao):
        print("z =",z)
        for k in SETUP_ITEMS[tao+z]: 
            print("k =",k)
 #### EXCEPTION 1
            if production.loc[k][tao] == 0:
                print("There is no setup in this period for product {}.".format(k))
                counter =+ 1
                continue
 #### EXCEPTION 2
            if demand.loc[k][tao+z] > spare_capacity[tao]['Spare Capacity']:
                print("Capacity in period {} is insufficient to pre-produce demands for product {} from period {}.\n".format(tao, k, tao+z))
                counter =+ 1
                continue
            if counter == k: 
                print("Stop Criterion is met!")
                break
##########################################################################
            if SM == 1:
                if SilverMeal(k,z) == True: 
                    print("Silver Meal Criterion is", SilverMeal(k,z))
                    production.loc[k][tao] += demand.loc[k][tao+z]
                    production.loc[k][tao+z] = 0
                else:
                    print("Else: Silver Meal Criterion is", SilverMeal(k,z))
            for t in range(tao,periods_count+1):
                for k in PRODUCTS:
                    spare_capacity[t] = capacity[t][1]-sum(production.loc[k][t] for k in PRODUCTS)
                SETUP_ITEMS = [[] for t in range(0,periods_count+1)]
                for t in PERIODS:
                    for k in PRODUCTS:
                        if production.loc[k][t]==(max(0,demand.loc[k][t]-stock.loc[k][t-1])) > 0: 
                            SETUP_ITEMS[t].append(k)   
            print(productionplan(production,spare_capacity,CF), '\n\n')
    print(productionplan(production,spare_capacity,CF), '\n\n')

The idea is, that if for one tao, there is an exception true for all k, all the loops terminate early, apart from the most outer one, so that we would go to the next tao in PERIODS and it all starts again.

I tried to use it with the counter variable, but this did not turn out to be functioning really well.

I currently have for example this output (extract):

z = 1
k = 1
Capacity in period 1 is insufficient to pre-produce demands for product 1 from period 2.

k = 2
Capacity in period 1 is insufficient to pre-produce demands for product 2 from period 2.

z = 2
k = 2
Capacity in period 1 is insufficient to pre-produce demands for product 2 from period 3.

After the k=2 in z=1the iteration should terminate, but it keeps on checking further z values.

Could anyone give me a tip how to solve this issue? I read about putting loops into functions, so that one can break out of multiple loops, but I am not sure how to formulate this here, as I would have multiple points of exit..

Thanks!

have a boolean variable terminateLoop. this variable is default set to false, when you want to terminate the loop, set the variable to true and break out of the loop. have an if condition at beginning of every nested loop, to check if the boolean is set to true, if yes then break. — abinmorth, Sep 12 '17 at 11:34
Hi @abinmorth. Thanks for your input. That helped me to fix it. :) good idea. — Frederic, Sep 15 '17 at 11:35

score 0 · Accepted Answer · answered Sep 12 '17 at 11:41

0

Python does not have control for breaking out of multiple loops at once.

You can set a flag and break out of multiple loops, for more info Link

answered Sep 12 '17 at 11:41

dalonlobo

484
4
18

Thank you Dalon! Helped a lot! – Frederic Sep 15 '17 at 11:36

Stop iterating over nested loops and continue with most outest loop

1 Answers1