how to stop a for loop

Question

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.

Answer 1

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;

Answer 2

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;
        }
    }
}   

}

Answer 3

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:

http://docs.python.org/tutorial/controlflow.html

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

Answer 4

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(any((cell != first for cell in col)) for elem in iterable)):

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

Answer 5

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)

Answer 6

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)