Find the first match

For example this very simple example in which we are looking for an even number:

examples/python/for_else/find_even_number.py

numbers = [3, 4, 5]

for n in numbers:
    if n % 2 == 0:
        even = n
        break

print(even)

If we run this program it will set the name even to be 4 which is what we wanted.

The problem is that if there is no match (there is no even number) then the variable even will never get a value and thus it will be undeclared.

examples/python/for_else/no_even_number.py

numbers = [3, 5]

for n in numbers:
    if n % 2 == 0:
        even = n
        break

print(even)

Running this program will result in the following exception:

$ python no_even_number.py

Traceback (most recent call last):
  File "no_even_number.py", line 8, in <module>
    print(even)
NameError: name 'even' is not defined

Default value outside the loop

One solution to this problem employed in most programming languages is to set a default value to the even variable before we enter the loop. That way if we pass all the values without finding a match, the variable even already has a value:

examples/python/for_else/default_result.py

numbers = [3, 5]

even = None
for n in numbers:
    if n % 2 == 0:
        even = n
        break

print(even)

$ python default_result.py

None

for-else

A more elegant way that can be written in Python is the use of else after the for-loop.

examples/python/for_else/for_else.py

numbers = [3, 5]

for n in numbers:
    if n % 2 == 0:
        even = n
        break
else:
    even = None

print(even)

The code in the else part will be executed if the loop has finished normally, without calling break.

That mean if we find a match during our loop, assign it to even and leave the loop calling break then the else part won't be executed.

Together with the name- or variable-scoping of Python this will result in the same effect, but in my humble opinion a more elegant way.