# 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
```python
# 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
```python
# 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
```python
# 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)
```
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"description": "\n# Break out of nested loops with max. 4 lines of code, no matter how deeply they are nested \n\n\n\n\n\n## The problem: breaking out of nested loops \n\n\n\n#### pip install break-out-nested \n\n\n\n```python\n\n# Break out of nested loops - a pain in the ***\n\n# How it is usually done:\n\ndone = False\n\nfor i in range(1, 6, 1): # 1st loop\n\n print('i:', i)\n\n for j in range(1, 11, 2): # 2nd loop\n\n print(' i, j:', i, j)\n\n for k in range(1, 21, 4): # 3rd loop\n\n print(' i,j,k:', i, j, k)\n\n if i % 3 == 0 and j % 3 == 0 and k % 3 == 0:\n\n done = True\n\n break # breaking from 3rd loop\n\n if done: break # breaking from 2nd loop\n\n if done: break # breaking from 1st loop\n\n```\n\n\n\n\n\n\n\n## The solution\n\n\n\n```python\n\n# Way easier\n\nfrom break_out_nested import it, bol\n\n\n\n# you need to create new variables as attributes of it,\n\n# because break_out_nested has only access to these variables\n\nit.i, it.j, it.k = 1, 1, 1\n\n\n\n\n\n# the break condition\n\ndef cond(): return it.i % 3 == 0 and it.j % 3 == 0 and it.k % 3 == 0\n\n\n\n\n\n# The condition will be checked in each loop\n\n# The function that checks the condition has to be passed as the last argument. \n\n# You can pass as many iterables as you want to the function\n\nfor it.i, it.j, it.k in bol(range(1, 6, 1), range(1, 11, 2), range(1, 21, 4), cond):\n\n print(it.i, it.j, it.k)\n\n```\n\n\n\n\n\n\n\n## More examples\n\n\n\n```python\n\n# More examples\n\ndef cond(): return it.i + it.j + it.k == 777\n\n\n\n\n\nit.i, it.j, it.k = 0, 0, 0\n\nfor it.i, it.j, it.k in bol(range(100), range(1000), range(10000), cond):\n\n print(it.i, it.j, it.k)\n\n\n\n\n\ndef cond(): return it.i + it.j + it.k >= 100000\n\n\n\n\n\nit.i, it.j, it.k = 0, 0, 0\n\n# you don't have to use it.i, it.j, it.k as the loop variables, you can\n\n# use anything you want, but you have to update the variables somewhere\n\nfor i, j, k in bol(range(100), range(1000), range(10000), cond):\n\n it.i, it.j, it.k = i * 10, j * 100, k * 100\n\n print(it.i, it.j, it.k)\n\n```\n\n\n\n",
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