how to break for loop in python

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 == "y" or ok == "Y": break 2 #this doesn't work :(         if ok == "n" or ok == "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?

Edit-FYI: get_input is a short function I wrote that supports showing a prompt and default values and all that fanciness and returns stdin.readline().strip()

Replay

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

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.

First, ordinary logic is helpful.

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.

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

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.

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.

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!

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.


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.

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...

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.

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.pop(x)
      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.pop(x)
      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.

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   # <--- !

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

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

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

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()
        if answer in "yn":
            return answer == "y"

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

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.

Category: python Time: 2008-10-10 Views: 4

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