# python – How to break out of multiple loops?

## The Question :

534 people think this question is useful

Given the following code (that doesn’t work):

while True:
#snip: print out current state
while True:
ok = get_input("Is this ok? (y/n)")
if ok.lower() == "y": break 2 #this doesn't work :(
if ok.lower() == "n": break
#do more processing with menus and stuff



Is there a way to make this work? Or do I have do one check to break out of the input loop, then another, more limited, check in the outside loop to break out all together if the user is satisfied?

563 people think this answer is useful

My first instinct would be to refactor the nested loop into a function and use return to break out.

306 people think this answer is useful

Here’s another approach that is short. The disadvantage is that you can only break the outer loop, but sometimes it’s exactly what you want.

for a in xrange(10):
for b in xrange(20):
if something(a, b):
# Break the inner loop...
break
else:
# Continue if the inner loop wasn't broken.
continue
# Inner loop was broken, break the outer.
break



This uses the for / else construct explained at: Why does python use ‘else’ after for and while loops?

Key insight: It only seems as if the outer loop always breaks. But if the inner loop doesn’t break, the outer loop won’t either.

The continue statement is the magic here. It’s in the for-else clause. By definition that happens if there’s no inner break. In that situation continue neatly circumvents the outer break.

154 people think this answer is useful

PEP 3136 proposes labeled break/continue. Guido rejected it because “code so complicated to require this feature is very rare”. The PEP does mention some workarounds, though (such as the exception technique), while Guido feels refactoring to use return will be simpler in most cases.

127 people think this answer is useful

If, for some reason, the terminating conditions can’t be worked out, exceptions are a fall-back plan.

class GetOutOfLoop( Exception ):
pass

try:
done= False
while not done:
isok= False
while not (done or isok):
ok = get_input("Is this ok? (y/n)")
if ok in ("y", "Y") or ok in ("n", "N") :
done= True # probably better
raise GetOutOfLoop
# other stuff
except GetOutOfLoop:
pass



For this specific example, an exception may not be necessary.

On other other hand, we often have “Y”, “N” and “Q” options in character-mode applications. For the “Q” option, we want an immediate exit. That’s more exceptional.

59 people think this answer is useful

I tend to agree that refactoring into a function is usually the best approach for this sort of situation, but for when you really need to break out of nested loops, here’s an interesting variant of the exception-raising approach that @S.Lott described. It uses Python’s with statement to make the exception raising look a bit nicer. Define a new context manager (you only have to do this once) with:

from contextlib import contextmanager
@contextmanager
def nested_break():
class NestedBreakException(Exception):
pass
try:
yield NestedBreakException
except NestedBreakException:
pass



Now you can use this context manager as follows:

with nested_break() as mylabel:
while True:
print "current state"
while True:
ok = raw_input("Is this ok? (y/n)")
if ok == "y" or ok == "Y": raise mylabel
if ok == "n" or ok == "N": break
print "more processing"



Advantages: (1) it’s slightly cleaner (no explicit try-except block), and (2) you get a custom-built Exception subclass for each use of nested_break; no need to declare your own Exception subclass each time.

43 people think this answer is useful

First, you may also consider making the process of getting and validating the input a function; within that function, you can just return the value if its correct, and keep spinning in the while loop if not. This essentially obviates the problem you solved, and can usually be applied in the more general case (breaking out of multiple loops). If you absolutely must keep this structure in your code, and really don’t want to deal with bookkeeping booleans…

You may also use goto in the following way (using an April Fools module from here):

#import the stuff
from goto import goto, label

while True:
#snip: print out current state
while True:
ok = get_input("Is this ok? (y/n)")
if ok == "y" or ok == "Y": goto .breakall
if ok == "n" or ok == "N": break
#do more processing with menus and stuff
label .breakall



I know, I know, “thou shalt not use goto” and all that, but it works well in strange cases like this.

40 people think this answer is useful

Introduce a new variable that you’ll use as a ‘loop breaker’. First assign something to it(False,0, etc.), and then, inside the outer loop, before you break from it, change the value to something else(True,1,…). Once the loop exits make the ‘parent’ loop check for that value. Let me demonstrate:

breaker = False #our mighty loop exiter!
while True:
while True:
if conditionMet:
#insert code here...
breaker = True
break
if breaker: # the interesting part!
break   # <--- !



If you have an infinite loop, this is the only way out; for other loops execution is really a lot faster. This also works if you have many nested loops. You can exit all, or just a few. Endless possibilities! Hope this helped!

27 people think this answer is useful

To break out of multiple nested loops, without refactoring into a function, make use of a “simulated goto statement” with the built-in StopIteration exception:

try:
for outer in range(100):
for inner in range(100):
if break_early():
raise StopIteration

except StopIteration: pass



See this discussion on the use of goto statements for breaking out of nested loops.

16 people think this answer is useful
keeplooping=True
while keeplooping:
#Do Stuff
while keeplooping:
#do some other stuff
if finisheddoingstuff():
keeplooping=False


or something like that. You could set a variable in the inner loop, and check it in the outer loop immediately after the inner loop exits, breaking if appropriate. I kinda like the GOTO method, provided you don’t mind using an April Fool’s joke module – its not Pythonic, but it does make sense.

13 people think this answer is useful

This isn’t the prettiest way to do it, but in my opinion, it’s the best way.

def loop():
while True:
#snip: print out current state
while True:
ok = get_input("Is this ok? (y/n)")
if ok == "y" or ok == "Y": return
if ok == "n" or ok == "N": break
#do more processing with menus and stuff



I’m pretty sure you could work out something using recursion here as well, but I dunno if that’s a good option for you.

11 people think this answer is useful

And why not to keep looping if two conditions are true? I think this is a more pythonic way:

dejaVu = True

while dejaVu:
while True:
ok = raw_input("Is this ok? (y/n)")
if ok == "y" or ok == "Y" or ok == "n" or ok == "N":
dejaVu = False
break



Isn’t it?

All the best.

8 people think this answer is useful

Factor your loop logic into an iterator that yields the loop variables and returns when done — here is a simple one that lays out images in rows/columns until we’re out of images or out of places to put them:

def it(rows, cols, images):
i = 0
for r in xrange(rows):
for c in xrange(cols):
if i >= len(images):
return
yield r, c, images[i]
i += 1

for r, c, image in it(rows=4, cols=4, images=['a.jpg', 'b.jpg', 'c.jpg']):
... do something with r, c, image ...



This has the advantage of splitting up the complicated loop logic and the processing…

4 people think this answer is useful

There is a hidden trick in the Python while ... else structure which can be used to simulate the double break without much code changes/additions. In essence if the while condition is false, the else block is triggered. Neither exceptions, continue or break trigger the else block. For more information see answers to “Else clause on Python while statement“, or Python doc on while (v2.7).

while True:
#snip: print out current state
ok = ""
while ok != "y" and ok != "n":
ok = get_input("Is this ok? (y/n)")
if ok == "n" or ok == "N":
break    # Breaks out of inner loop, skipping else

else:
break        # Breaks out of outer loop

#do more processing with menus and stuff



The only downside is that you need to move the double breaking condition into the while condition (or add a flag variable). Variations of this exists also for the for loop, where the else block is triggered after loop completion.

3 people think this answer is useful

In this case, as pointed out by others as well, functional decomposition is the way to go. Code in Python 3:

def user_confirms():
while True:
answer = input("Is this OK? (y/n) ").strip().lower()

def main():
while True:
# do stuff
if user_confirms():
break



3 people think this answer is useful

By using a function:

def myloop():
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:
return  # getting out of all loops

myloop()



Try running the above codes by commenting out the return as well.

Without using any function:

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



Now, run the above codes as is first and then try running by commenting out each line containing break one at a time from the bottom.

3 people think this answer is useful

An easy way to turn multiple loops into a single, breakable loop is to use numpy.ndindex

for i in range(n):
for j in range(n):
val = x[i, j]
break # still inside the outer loop!

for i, j in np.ndindex(n, n):
val = x[i, j]
break # you left the only loop there was!



You do have to index into your objects, as opposed to being able to iterate through the values explicitly, but at least in simple cases it seems to be approximately 2-20 times simpler than most of the answers suggested.

2 people think this answer is useful

Another way of reducing your iteration to a single-level loop would be via the use of generators as also specified in the python reference

for i, j in ((i, j) for i in A for j in B):
print(i , j)
if (some_condition):
break



You could scale it up to any number of levels for the loop

The downside is that you can no longer break only a single level. It’s all or nothing.

Another downside is that it doesn’t work with a while loop. I originally wanted to post this answer on Python – break out of all loops but unfortunately that’s closed as a duplicate of this one

2 people think this answer is useful

My reason for coming here is that i had an outer loop and an inner loop like so:

for x in array:
for y in dont_use_these_values:
if x.value==y:
array.remove(x)  # fixed, was array.pop(x) in my original answer
continue

do some other stuff with x



As you can see, it won’t actually go to the next x, but will go to the next y instead.

what i found to solve this simply was to run through the array twice instead:

for x in array:
for y in dont_use_these_values:
if x.value==y:
array.remove(x)  # fixed, was array.pop(x) in my original answer
continue

for x in array:
do some other stuff with x



I know this was a specific case of OP’s question, but I am posting it in the hope that it will help someone think about their problem differently while keeping things simple.

2 people think this answer is useful

I’d like to remind you that functions in Python can be created right in the middle of the code and can access the surrounding variables transparently for reading and with nonlocal or global declaration for writing.

So you can use a function as a “breakable control structure”, defining a place you want to return to:

def is_prime(number):

foo = bar = number

def return_here():
nonlocal foo, bar
init_bar = bar
while foo > 0:
bar = init_bar
while bar >= foo:
if foo*bar == number:
return
bar -= 1
foo -= 1

return_here()

if foo == 1:
print(number, 'is prime')
else:
print(number, '=', bar, '*', foo)



>>> is_prime(67)
67 is prime
>>> is_prime(117)
117 = 13 * 9
>>> is_prime(16)
16 = 4 * 4



2 people think this answer is useful

Try using an infinite generator.

from itertools import repeat
inputs = (get_input("Is this ok? (y/n)") for _ in repeat(None))
response = (i.lower()=="y" for i in inputs if i.lower() in ("y", "n"))

while True:
#snip: print out current state
if next(response):
break
#do more processing with menus and stuff



2 people think this answer is useful
# this version uses a level counter to choose how far to break out

break_levels = 0
while True:
# snip: print out current state
while True:
ok = get_input("Is this ok? (y/n)")
if ok == "y" or ok == "Y":
break_levels = 1        # how far nested, excluding this break
break
if ok == "n" or ok == "N":
break                   # normal break
if break_levels:
break_levels -= 1
break                       # pop another level
if break_levels:
break_levels -= 1
break

# ...and so on



2 people think this answer is useful

There is no way to do this from a language level. Some languages have a goto others have a break that takes an argument, python does not.

The best options are:

1. Set a flag which is checked by the outer loop, or set the outer loops condition.

2. Put the loop in a function and use return to break out of all the loops at once.

Credit goes to Vivek Nagarajan, Programmer since 1987

Using Function

def doMywork(data):
for i in data:
for e in i:
return



Using flag

is_break = False
for i in data:
if is_break:
break # outer loop break
for e in i:
is_break = True
break # inner loop break



1 people think this answer is useful

probably little trick like below will do if not prefer to refactorial into function

added 1 break_level variable to control the while loop condition

break_level = 0
# while break_level < 3: # if we have another level of nested loop here
while break_level < 2:
#snip: print out current state
while break_level < 1:
ok = get_input("Is this ok? (y/n)")
if ok == "y" or ok == "Y": break_level = 2 # break 2 level
if ok == "n" or ok == "N": break_level = 1 # break 1 level



1 people think this answer is useful

You can define a variable( for example break_statement ), then change it to a different value when two-break condition occurs and use it in if statement to break from second loop also.

while True:
break_statement=0
while True:
ok = raw_input("Is this ok? (y/n)")
if ok == "n" or ok == "N":
break
if ok == "y" or ok == "Y":
break_statement=1
break
if break_statement==1:
break



1 people think this answer is useful

# Solutions in 2 Ways

With an example: Are these two matrices equal/same?
matrix1 and matrix2 are same size, n, 2 dimentional matrices.

First Solution, without a function

same_matrices = True
inner_loop_broken_once = False
n = len(matrix1)

for i in range(n):
for j in range(n):

if matrix1[i][j] != matrix2[i][j]:
same_matrices = False
inner_loop_broken_once = True
break

if inner_loop_broken_once:
break



Second Solution, with a function
This is the final solution for my case

def are_two_matrices_the_same (matrix1, matrix2):
n = len(matrix1)
for i in range(n):
for j in range(n):
if matrix1[i][j] != matrix2[i][j]:
return False
return True



Have a nice day!

1 people think this answer is useful
# this version breaks up to a certain label

break_label = None
while True:
# snip: print out current state
while True:
ok = get_input("Is this ok? (y/n)")
if ok == "y" or ok == "Y":
break_label = "outer"   # specify label to break to
break
if ok == "n" or ok == "N":
break
if break_label:
if break_label != "inner":
break                   # propagate up
break_label = None          # we have arrived!
if break_label:
if break_label != "outer":
break                       # propagate up
break_label = None              # we have arrived!

#do more processing with menus and stuff



0 people think this answer is useful

Hopefully this helps:

x = True
y = True
while x == True:
while y == True:
ok = get_input("Is this ok? (y/n)")
if ok == "y" or ok == "Y":
x,y = False,False #breaks from both loops
if ok == "n" or ok == "N":
break #breaks from just one



0 people think this answer is useful

Here’s an implementation that seems to work:

break_ = False
for i in range(10):
if break_:
break
for j in range(10):
if j == 3:
break_ = True
break
else:
print(i, j)



The only draw back is that you have to define break_ before the loops.

-3 people think this answer is useful

Similar like the one before, but more compact. (Booleans are just numbers)

breaker = False #our mighty loop exiter!
while True:
while True:
ok = get_input("Is this ok? (y/n)")
breaker+= (ok.lower() == "y")
break

if breaker: # the interesting part!
break   # <--- !



-3 people think this answer is useful

Since this question has become a standard question for breaking into a particular loop, I would like to give my answer with example using Exception.

Although there exists no label named breaking of loop in multipally looped construct, we can make use of User-defined Exceptions to break into a particular loop of our choice. Consider the following example where let us print all numbers upto 4 digits in base-6 numbering system:

class BreakLoop(Exception):
def __init__(self, counter):
Exception.__init__(self, 'Exception 1')
self.counter = counter

for counter1 in range(6):   # Make it 1000
try:
thousand = counter1 * 1000
for counter2 in range(6):  # Make it 100
try:
hundred = counter2 * 100
for counter3 in range(6): # Make it 10
try:
ten = counter3 * 10
for counter4 in range(6):
try:
unit = counter4
value = thousand + hundred + ten + unit
if unit == 4 :
raise BreakLoop(4) # Don't break from loop
if ten == 30:
raise BreakLoop(3) # Break into loop 3
if hundred == 500:
raise BreakLoop(2) # Break into loop 2
if thousand == 2000:
raise BreakLoop(1) # Break into loop 1

print('{:04d}'.format(value))
except BreakLoop as bl:
if bl.counter != 4:
raise bl
except BreakLoop as bl:
if bl.counter != 3:
raise bl
except BreakLoop as bl:
if bl.counter != 2:
raise bl
except BreakLoop as bl:
pass



When we print the output, we will never get any value whose unit place is with 4. In that case, we don’t break from any loop as BreakLoop(4) is raised and caught in same loop. Similarly, whenever ten place is having 3, we break into third loop using BreakLoop(3). Whenever hundred place is having 5, we break into second loop using BreakLoop(2) and whenver the thousand place is having 2, we break into first loop using BreakLoop(1).

In short, raise your Exception (in-built or user defined) in the inner loops, and catch it in the loop from where you want to resume your control to. If you want to break from all loops, catch the Exception outside all the loops. (I have not shown this case in example).