while i < 5:
do stuff
if i == 3:
print "i is 3"
continue
Top answer 1 of 2
13
while i < 5:
do stuff
if i == 3:
print "i is 3"
continue
2 of 2
2
Instead of break use continue
Now, I pretty much never use continue as I find it is usually clearer to rework the code to avoid it. Of course that's really easy in this example, if you have trouble with a more complex example ask about that one.
freeCodeCamp
freecodecamp.org › news › break-in-python-nested-for-loop-break-if-condition-met-example
Break in Python – Nested For Loop Break if Condition Met Example
May 17, 2022 - The example in this section will be similar to the last section's. We'll be using a while loop instead. ... The code above prints a range of numbers from 0 to 9. We're going to use the break to stop printing numbers when we get to 5. i = 1 while i < 10: print(i) if i == 5: break i += 1 · Just like we did in the last section, we created a new condition: if i == 5 and when this condition is met, the loop is terminated instead of printing all the way to 9.
python - How to stop one or multiple for loop(s) - Stack Overflow
I find it easier to understand with the use of a while loop and it will stop both for loops that way. The code below also return the True/False as asked when the function check_nxn_list() is called. More on stackoverflow.com
Ending a "For" loop when condition is not met
you have a lot of unnecessary code in your function imo. all you need to check is that it has a vowel. if you loop through the word and there are no vowels, you can just exit after the first pass. no need for the other two conditions. #My best attempt def has_a_vowel(a_str): #If a_str is empty it will return False if a_str == "": return False #Review every letter in a_str for letter in a_str: #If letter is a vowel, return the value True if letter in "aeiou": return True return False print(has_a_vowel("abba")) print(has_a_vowel("beeswax")) print(has_a_vowel("syzygy")) print(has_a_vowel("")) will return True True False False More on reddit.com
8
1
November 30, 2020python - Stop for loop or while loop when condition met - Stack Overflow
I am newby in programming. The goal is to count numbers in list in order, but loop has to stop when condition is met or close to it, but must not exceed it. For example: list = [4,4,4,3,3], conditi... More on stackoverflow.com
python - Stopping a function with a while loop when a certain condition is met - Stack Overflow
I basically want to know how I can stop the while-loop from adding to the list when it reaches a certain element of the list ("LA Dodgers Stadium") in this case. It should not add anything after it reaches that element of the list. I know I asked this question before but I did not recieve a proper answer. ... You might wish to alter you loop condition... More on stackoverflow.com
DigitalOcean
digitalocean.com › community › tutorials › how-to-use-break-continue-and-pass-statements-when-working-with-loops-in-python-3
How to Break Out of Multiple Loops in Python | DigitalOcean
August 7, 2025 - Wrapping your loop logic in a function and using return to exit is a Pythonic and efficient method to stop execution when a condition is met.
Coursera
coursera.org › tutorials › python-break
How to Use Python Break | Coursera
In Python, break allows you to exit a loop when an external condition is met. Normal program execution resumes at the next statement. You can use a break statement with both for loops and while loops.
Top answer 1 of 7
129
Use break and continue to do this. Breaking nested loops can be done in Python using the following:
for a in range(...):
for b in range(..):
if some condition:
# break the inner loop
break
else:
# will be called if the previous loop did not end with a `break`
continue
# but here we end up right after breaking the inner loop, so we can
# simply break the outer loop as well
break
Another way is to wrap everything in a function and use return to escape from the loop.
2 of 7
72
There are several ways to do it:
The simple Way: a sentinel variable
n = L[0][0]
m = len(A)
found = False
for i in range(m):
if found:
break
for j in range(m):
if L[i][j] != n:
found = True
break
Pros: easy to understand Cons: additional conditional statement for every loop
The hacky Way: raising an exception
n = L[0][0]
m = len(A)
try:
for x in range(3):
for z in range(3):
if L[i][j] != n:
raise StopIteration
except StopIteration:
pass
Pros: very straightforward Cons: you use Exception outside of their semantic
The clean Way: make a function
def is_different_value(l, elem, size):
for x in range(size):
for z in range(size):
if l[i][j] != elem:
return True
return False
if is_different_value(L, L[0][0], len(A)):
print "Doh"
pros: much cleaner and still efficient cons: yet feels like C
The pythonic way: use iteration as it should be
def is_different_value(iterable):
first = iterable[0][0]
for l in iterable:
for elem in l:
if elem != first:
return True
return False
if is_different_value(L):
print "Doh"
pros: still clean and efficient cons: you reinvdent the wheel
The guru way: use any():
def is_different_value(iterable):
first = iterable[0][0]
return any(cell != first for col in iterable for cell in col)
if is_different_value(L):
print "Doh"
pros: you'll feel empowered with dark powers cons: people that will read you code may start to dislike you
Reddit
reddit.com › r/learnpython › ending a "for" loop when condition is not met
r/learnpython on Reddit: Ending a "For" loop when condition is not met
November 30, 2020 -
Edit: SOLVED
I was going to post this in the Ask Anything Weekly Monday Thread but it seemed a little long so here goes.
I am trying to end a "for" loop if the condition is not met. Specifically, I need to write a function that will check if a string contains any vowels, if so return True, if not return False. I am struggling with getting my function to return False when the string does not contain any vowels. Right now, my function always returns None if there are no vowels.
The correct answer should return:
TrueTrueFalseFalse
My best answer returns:
TrueTrueNoneFalse
I believe I am getting a "None" because nothing is being returned when the function is run and the string contains no vowels. I think the "continue" loop in Line 13 and 14 are the reason. Or maybe that's not the problem.... I'm so stuck.
I have included the quiz, along with my best attempt at answering it below
EDIT: A big thank you to everyone for your help and explanations!!
#My best attempt
def has_a_vowel(a_str):
#If a_str is empty it will return False
if a_str == "":
return False
#Review every letter in a_str
for letter in a_str:
#If letter is a vowel, return the value True
if letter in "aeiou":
return True
#If letter is not a vowel, continue checking the other letters
elif letter not in "aeiou":
continue
#Here I want to say if there are no vowels in a_str, return false
else:
if letter not in "aeiou":
return False
print(has_a_vowel("abba"))
print(has_a_vowel("beeswax"))
print(has_a_vowel("syzygy"))
print(has_a_vowel("")) Top answer 1 of 3
3
you have a lot of unnecessary code in your function imo. all you need to check is that it has a vowel. if you loop through the word and there are no vowels, you can just exit after the first pass. no need for the other two conditions. #My best attempt def has_a_vowel(a_str): #If a_str is empty it will return False if a_str == "": return False #Review every letter in a_str for letter in a_str: #If letter is a vowel, return the value True if letter in "aeiou": return True return False print(has_a_vowel("abba")) print(has_a_vowel("beeswax")) print(has_a_vowel("syzygy")) print(has_a_vowel("")) will return True True False False
2 of 3
2
Seems like a great use case for a while loop!
IONOS
ionos.com › digital guide › websites › web development › python break continue
How to use Python break and continue - IONOS
July 13, 2023 - A Python break which says that the loop should break when the number 5 is reached can then be inserted. 5 is within the specified range, meaning that the loop will terminate and the code will continue after that. It will look like this: for num in range(10): if num == 5: print("The termination condition is met") break print(f"The current number is {num}") print("Continuation after loop")
Real Python
realpython.com › python-while-loop
Python while Loops: Repeating Tasks Conditionally – Real Python
March 3, 2025 - The next script is almost identical to the one in the previous section except for the continue statement in place of break: ... This time, when number is 2, the continue statement terminates that iteration.
Stack Overflow
stackoverflow.com › questions › 69722136 › stop-for-loop-or-while-loop-when-condition-met
python - Stop for loop or while loop when condition met - Stack Overflow
October 26, 2021 - Use another variable name like l. Then, if you run the loop in a function, it is simpler to just return from inside the loop if the condition is met ... Save this answer. Show activity on this post.
freeCodeCamp
freecodecamp.org › news › python-break-statement-tutorial
Python Break Statement – How to Break Out of a For Loop in Python
April 16, 2024 - Once again, "Jade" and "John" were printed because the loop stops when "John" is found. In this article, you learned to use the break statement in Python. You can use it to terminate the current loop when a condition is met.
Top answer 1 of 5
4
You might wish to alter you loop condition. Instead of while True try:
place = places[0]
while place != "LA Dodgers Stadium" and count < len(places):
if ' ' in place:
multi_word += 1
count += 1
place = places[count]
Edit: possibly a better way of writing this might be:
for place in places:
if place == "LA Dodgers Stadium": break
if ' ' in place: multi_word += 1
2 of 5
0
(pseudo code)
flag = true
while (flag)
...
if (place == LA Dodgers stadium)
{
flag = false
}
Enki
enki.com › post › how-to-exit-a-loop-in-python
Enki | Blog - How to Exit a Loop in Python
Learn to manage loops effectively in Python using the break statement to exit for and while loops under specific conditions, optimizing code efficiency.
Spark By {Examples}
sparkbyexamples.com › home › python › python for loop continue and break
Python For Loop Continue And Break - Spark By {Examples}
April 17, 2024 - Note: continue statement doesn’t exit the for/while loop. ... Sometimes you would like to exit from the python for/while loop when you meet certain conditions, using the break statement you can exit the loop when the condition meets.
Programiz
programiz.com › python-programming › break-continue
Python break and continue (With Examples)
Try Programiz PRO! ... The break statement terminates the loop immediately when it's encountered. ... The above image shows the working of break statements in for and while loops. Note: The break statement is usually used inside decision-making statements such as if...else.
Manifoldapp
cuny.manifoldapp.org › read › how-to-code-in-python-3 › section › 8adf4bec-a6ca-4a11-8c8d-4cf4052d5ac4
How To Use Break, Continue, and Pass Statements when Working with Loops | How To Code in Python 3 | Manifold @CUNY
In Python, the break statement provides you with the opportunity to exit out of a loop when an external condition is triggered.
Replit
replit.com › home › discover › how to stop a loop in python
How to stop a loop in Python | Replit
February 13, 2026 - Once the condition is met, the flag is set to False. The loop then terminates gracefully after completing its current iteration. This approach offers a "soft" exit, allowing the rest of the code in the current iteration to run before stopping, ...
Top answer 1 of 4
6
A for loop ends when it exhausts its iterable, and a while loop terminates when its condition is no longer met. If you were going to write while True: if some_variable == "some value": break #do stuff then instead you could write: while some_variable != "some value": #do stuff
2 of 4
1
See itertools takewhile and dropwhile.
PsychoPy
discourse.psychopy.org › builder
How to break a loop if a certain condition is met - Builder - PsychoPy
December 3, 2022 - What are you trying to achieve?: So I have a nested loop (the picture above) in my experiment. The inner loop has three conditions. I wanna be able to skip the other 2 or 1 condition when the correct response is recei…