Several nested 'for' loops, continue to next iteration of outer loop if condition inside inner loop is true [duplicate]

Question

I know it is terribly inefficient and ugly code, but if I have three for loops, nested inside each other such as so:

for x in range(0, 10):
    for y in range(x+1, 11):
       for z in range(y+1, 11):
           if ...

I want to break the two inner loops and continue to the next iteration of the outer loop if the if statement is true. Can this be done?

Constantino

AFAIK the Python founder refused to add a construct to do that because the code would become ugly – Twotime

you could write it using variables to store state and test everytime you break / finish the loop – Brouwer

@wil93 Are you referring to the constructs break and continue? They are recognised in Python, well in Python 3.4 anyway. – Constantino

@Brouwer Yes, that does seem like the neatest solution possible, as tglaria has demostrated in his answer below. – Constantino

I was referring to the labeled break/continue construct, but as I said it was rejected. – Twotime

Answer 1

Check some variable after each loops ends:

for x in range(0, 10):
    for y in range(x+1, 11):
        for z in range(y+1, 11):
            if condition:
                variable = True
                break
            #...
        if variable:
            break;
        #...

Marismarisa

Answer 2

Another option is to use exceptions instead of state variables:

class BreakException(Exception):
    pass

for x in range(0, 10):
    try:
        for y in range(x+1, 11):
           for z in range(y+1, 11):
               if True:
                   raise BreakException
    except BreakException:
        pass

I imagine this could be especially useful if bailing out of more than two inner loops.

Knelt

Very helpful. This should be marked as the answer! – Lanner

This is an anti-pattern, exceptions should not be used for the logic side of the code – Laryngitis

@Laryngitis tell that to the Python Devs who thought StopIteration was a good idea. Anyway, you are right. There are cleaner solutions, and I'm just realizing one of them has not been posted yet... – Knelt

Answer 3

n = False
for x in range(0,10):
    if n == True:
        print(x,y,z)
    for y in range(x+1, 11):
        if n == True:
            break
        for z in range(y+1, 11):
            if z == 5:
                n = True
                break

(1, 2, 5)
(2, 2, 5)
(3, 3, 5)
(4, 4, 5)
(5, 5, 5)
(6, 6, 5)
(7, 7, 5)
(8, 8, 5)
(9, 9, 5)

Brouwer

Answer 4

A possible solution is to merge the two inner loops to a single one (that can be terminated with break):

import itertools

for x in range(10):
    for y, z in itertools.combinations(range(x+1, 11), 2):
        if condition:
            break

Twotime

Answer 5

Define a function for the inner loops and simply return when the condition is True.

def process_inner(x):
    for y in range(x+1, 11):
       for z in range(y+1, 11):
           if condition:
               return
           # ...


for x in range(0, 10):
    process_inner(x)

I now believe this is a cleaner solution than abusing exceptions or introducing control flow variables.

It's not obvious in the abstract example, but the fact that some loops should be aborted and others not is a sign that these loops have very different purposes. When this is the case, factoring them into separate functions will communicate that difference much clearer than a bunch of nested loops.

Knelt