Breaking out of nested loops

PythonLoopsNested Loops

Python Problem Overview


Is there an easier way to break out of nested loops than throwing an exception? (In Perl, you can give labels to each loop and at least continue an outer loop.)

for x in range(10):
    for y in range(10):
        print x*y
        if x*y > 50:
            "break both loops"

I.e., is there a nicer way than:

class BreakIt(Exception): pass

try:
    for x in range(10):
        for y in range(10):
            print x*y
            if x*y > 50:
                raise BreakIt
except BreakIt:
    pass

Python Solutions


Solution 1 - Python

for x in xrange(10):
    for y in xrange(10):
        print x*y
        if x*y > 50:
            break
    else:
        continue  # only executed if the inner loop did NOT break
    break  # only executed if the inner loop DID break

The same works for deeper loops:

for x in xrange(10):
    for y in xrange(10):
        for z in xrange(10):
            print x,y,z
            if x*y*z == 30:
                break
        else:
            continue
        break
    else:
        continue
    break

Solution 2 - Python

It has at least been suggested, but also rejected. I don't think there is another way, short of repeating the test or re-organizing the code. It is sometimes a bit annoying.

In the rejection message, Mr van Rossum mentions using return, which is really sensible and something I need to remember personally. :)

Solution 3 - Python

If you're able to extract the loop code into a function, a return statement can be used to exit the outermost loop at any time.

def foo():
    for x in range(10):
        for y in range(10):
            print(x*y)
            if x*y > 50:
                return
foo()

If it's hard to extract that function you could use an inner function, as @bjd2385 suggests, e.g.

def your_outer_func():
    ...
    def inner_func():
        for x in range(10):
            for y in range(10):
                print(x*y)
                if x*y > 50:
                    return
    inner_func()
    ...

Solution 4 - Python

Use itertools.product!

from itertools import product
for x, y in product(range(10), range(10)):
    #do whatever you want
    break

Here's a link to itertools.product in the python documentation: http://docs.python.org/library/itertools.html#itertools.product

You can also loop over an array comprehension with 2 fors in it, and break whenever you want to.

>>> [(x, y) for y in ['y1', 'y2'] for x in ['x1', 'x2']]
[    ('x1', 'y1'), ('x2', 'y1'),    ('x1', 'y2'), ('x2', 'y2')]

Solution 5 - Python

Sometimes I use a boolean variable. Naive, if you want, but I find it quite flexible and comfortable to read. Testing a variable may avoid testing again complex conditions and may also collect results from several tests in inner loops.

    x_loop_must_break = False
    for x in range(10):
        for y in range(10):
            print x*y
            if x*y > 50:
                x_loop_must_break = True
                break
        if x_loop_must_break: break

Solution 6 - Python

If you're going to raise an exception, you might raise a StopIteration exception. That will at least make the intent obvious.

Solution 7 - Python

You can also refactor your code to use a generator. But this may not be a solution for all types of nested loops.

Solution 8 - Python

In this particular case, you can merge the loops with a modern python (3.0 and probably 2.6, too) by using itertools.product.

I for myself took this as a rule of thumb, if you nest too many loops (as in, more than 2), you are usually able to extract one of the loops into a different method or merge the loops into one, as in this case.

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionMichael KuhnView Question on Stackoverflow
Solution 1 - PythonMarkus JarderotView Answer on Stackoverflow
Solution 2 - PythonunwindView Answer on Stackoverflow
Solution 3 - PythonMr FoozView Answer on Stackoverflow
Solution 4 - PythonFábio SantosView Answer on Stackoverflow
Solution 5 - PythonHalberdierView Answer on Stackoverflow
Solution 6 - PythondbnView Answer on Stackoverflow
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