I'm writing a code to determine if every element in my nxn list is the same. i.e. [[0,0],[0,0]] returns true but [[0,1],[0,0]] will return false. I was thinking of writing a code that stops immediately when it finds an element that is not the same as the first element. i.e:

n=L[0][0] m=len(A) for i in range(m): for j in range(m): if L[i][j]==n: -continue the loop- else: -stop the loop- 

I would like to stop this loop if L[i][j]!==n and return false. otherwise return true. How would I go about implementing this?

9 Answers

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.

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

Try to simply use break statement.

Also you can use the following code as an example:

a = [[0,1,0], [1,0,0], [1,1,1]] b = [[0,0,0], [0,0,0], [0,0,0]] def check_matr(matr, expVal): for row in matr: if len(set(row)) > 1 or set(row).pop() != expVal: print 'Wrong' break# or return else: print 'ok' else: print 'empty' check_matr(a, 0) check_matr(b, 0) 

Use the break statement:

Others ways to do the same is:

el = L[0][0] m=len(L) print L == [[el]*m]*m 

Or:

first_el = L[0][0] print all(el == first_el for inner_list in L for el in inner_list) 

To achieve this you would do something like:

n=L[0][0] m=len(A) for i in range(m): for j in range(m): if L[i][j]==n: //do some processing else: break; 

In order to jump out of a loop, you need to use the break statement.

n=L[0][0] m=len(A) for i in range(m): for j in range(m): if L[i][j]!=n: break; 

Here you have the official Python manual with the explanation about break and continue, and other flow control statements:

EDITED: As a commenter pointed out, this does only end the inner loop. If you need to terminate both loops, there is no "easy" way (others have given you a few solutions). One possiblity would be to raise an exception:

def f(L, A): try: n=L[0][0] m=len(A) for i in range(m): for j in range(m): if L[i][j]!=n: raise RuntimeError( "Not equal" ) return True except: return False 

I know this question was asked a long time ago, but if that could help anyone else, here's my answer:

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. Some parameters may have to be added in the function call.

def check_nxn_list(): state = True n=L[0][0] m=len(A) while state == True: for i in range(m): for j in range(m): if L[i][j]!=n: state = False break return state 

The break at the end of the while loop is required to end the loop even if state stays True.

To stop your loop you can use break with label. It will stop your loop for sure. Code is written in Java but aproach is the same for the all languages.

public void exitFromTheLoop() { boolean value = true; loop_label:for (int i = 0; i < 10; i++) { if(!value) { System.out.println("iteration: " + i); break loop_label; } } } 

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