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=1 the 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!

Answer 1

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