tags:

views:

1077

answers:

6

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
+13  A: 

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. :)

unwind
+1  A: 

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.

Tetha
+2  A: 

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

J S
A: 

At risk of being stoned to death - http://entrian.com/goto/download.html

Ford_Prefect
*throws first stone*
Epaga
Christian Witts
+3  A: 
for x in xrange(10):
    for y in xrange(10):
        print x*y
        if x*y > 50:
            break
    else:
        continue  # only executed if break was not used
    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
MizardX
+6  A: 
def foo():
    for x in range(10):
        for y in range(10):
            print x*y
            if x*y > 50:
                return
foo()
Mr Fooz