Python For If Else

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 t

Solution 1:

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 = [xforx in lst ifx%2 == 0]
if filtered:
    for item in filtered:
        print item
else:
    print"No evens"

Solution 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
        breakelse:
    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
        breakelse:
    print"There are no EVEN #s"

Solution 3:

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

e.g.

flag = Falsefor i in [0,2,4]:
    if i%2==0:
        print i
        flag = Trueif not flag:
    print"There is an ODD #"

Solution 4:

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

ifany(i % 2 == 0for 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.