break-out-nested 0.10

Break out of nested loops with max. 4 lines of code, no matter how deeply they are nested

Break out of nested loops with max. 4 lines of code, no matter how deeply they are nested

The problem: breaking out of nested loops

pip install break-out-nested

# Break out of nested loops - a pain in the ***

# How it is usually done:

done = False

for i in range(1, 6, 1):  # 1st loop

    print('i:', i)

    for j in range(1, 11, 2):  # 2nd loop

        print('   i, j:', i, j)

        for k in range(1, 21, 4):  # 3rd loop

            print('      i,j,k:', i, j, k)

            if i % 3 == 0 and j % 3 == 0 and k % 3 == 0:

                done = True

                break  # breaking from 3rd loop

        if done: break  # breaking from 2nd loop

    if done: break  # breaking from 1st loop

The solution

# Way easier

from break_out_nested import it, bol



# you need to create new variables as attributes of it,

# because break_out_nested has only access to these variables

it.i, it.j, it.k = 1, 1, 1





# the break condition

def cond(): return it.i % 3 == 0 and it.j % 3 == 0 and it.k % 3 == 0





# The condition will be checked in each loop

# The function that checks the condition has to be passed as the last argument. 

# You can pass as many iterables as you want to the function

for it.i, it.j, it.k in bol(range(1, 6, 1), range(1, 11, 2), range(1, 21, 4), cond):

    print(it.i, it.j, it.k)

More examples

# More examples

def cond(): return it.i + it.j + it.k == 777





it.i, it.j, it.k = 0, 0, 0

for it.i, it.j, it.k in bol(range(100), range(1000), range(10000), cond):

    print(it.i, it.j, it.k)





def cond(): return it.i + it.j + it.k >= 100000





it.i, it.j, it.k = 0, 0, 0

# you don't have to use it.i, it.j, it.k as the loop variables, you can

# use anything you want, but you have to update the variables somewhere

for i, j, k in bol(range(100), range(1000), range(10000), cond):

    it.i, it.j, it.k = i * 10, j * 100, k * 100

    print(it.i, it.j, it.k)