1

I just realized that this code does not work the way I hoped it would worked...

for i in [0,2,4]:
    if i%2==0:
        print i
else:
    print "There are no EVEN #s"

So what's the best way of coding this sort of structure

The idea is to run the "else" statement code if the "if" statement never becomes true through all the iterations of the "for" loop

Riddler
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    If you never `break` (or `return`), the `else` always runs. – jonrsharpe Aug 14 '14 at 22:53
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    In order to avoid the `else` statement in a `for` loop you need to `break` the loop. You'd need a state variable that started as `False`, was set to `True` in your conditional block, and then a conditional `break`. – g.d.d.c Aug 14 '14 at 22:53
  • I ve used that logic before, but just out of curiosity is there another way of using try,except, catch? or something – Riddler Aug 14 '14 at 22:55
  • @Riddler just add a `break` when you find the first even number. There are other ways of doing this, of course. Also, it would be helpful to *actually include* how you hoped the code would work, and how it works instead. – jonrsharpe Aug 14 '14 at 22:58
  • You'll need to implement your logic. One way will be to count even numbers first, and check count after the loop. Or you can set flag when even number is found. – Ivan Nevostruev Aug 14 '14 at 22:58

4 Answers4

2

The code doesn't work as you would like it to because the if and else are not on the same level of scope. However, there is a for...else syntax in Python that you were perhaps trying to use. For information on that, see here. To use the for...else syntax, you need to have a break statement inside the for loop. If it breaks, then the else is not called, otherwise the else be called after the loop is done.

However, if you don't have a break statement, then else always runs.

Here is your code, corrected:

for i in [0,2,4]:
    if i%2==0:
        print i
        break
else:
    print "There are no EVEN #s"

As soon as the loop encounters an even number, the loop breaks. Otherwise, if the loop would fully execute (i.e. go through the entire list), then it would also run the else. Just for reference, here is the loop on a list of odd numbers:

for i in [1,3,5]:
    if i%2==0:
        print i
        break
else:
    print "There are no EVEN #s"
Community
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Newb
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2

I think that a more "pythonic" approach would be to filter the list and check if the filtered list has any elements to figure out what to print:

lst = [0, 2, 4]
filtered = [x for x in lst if x%2 == 0]
if filtered:
    for item in filtered:
        print item
else:
    print "No evens"
mgilson
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1

just set a flag in the if statement that you can check after the for loop is finished running.

e.g.

flag = False
for i in [0,2,4]:
    if i%2==0:
        print i
        flag = True

if not flag:
    print "There is an ODD #"
bgenchel
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    It'd probably be more pythonic to let `flag = True` in the start, and in the `if` statement in the loop set `flag = False`, so then in the final `if` statement you can have `if flag` instead of `if not flag`. – Newb Aug 14 '14 at 23:07
0

If you just want to check if any even numbers exist and don't care about the value use any:

if any(i % 2 == 0 for i in [0,2,4,5]):
        print "The list has even numbers"          
else:
    print "There are no EVEN #s"

any will evaluate lazily returning True as soon as any even number is found or False if the list contains no even numbers.

Padraic Cunningham
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