Print all even numbers in a list until a given number

Question

I am just beginning to dabble in Python, and have started to go through the chapters on learnpython.org. In the \'Loops\' chapter, I have solved the challenge with the follo

Answer 1

3 if's and done :)

    if x % 2 == 0:
        print x
    if x == 237:
        break
    if x % 2 != 0: 
        continue

Output

402
984
360
408
980
544
390
984
592
236
942
386
462
418
344
236
566
978
328
162
758
918

Answer 2

Im just going throug this exercise in my "learning the basis thing" and i came out with a similar but somehow worse solution, patrik :(

x = 0 for loop in numbers: if (numbers[x]) % 2 ==0: print (numbers[x]) x = x+1
if (numbers[x]) ==237: break
else: x =x+1 continue

Answer 3

That's what else and elif are for:

for x in numbers:
  if x == 237:
    break
  elif x % 2 == 0:
    print x

Answer 4

This is also possible:

try:
    i = numbers.index(237)
except:
    i = len(numbers)
for n in numbers[:i]:
    if not n%2:
       print n

Answer 5

you can do that using list comprehension

The solution would be like this:

numbers = [x for ind,x in enumerate(numbers) if x % 2== 0 and numbers[ind+1]!=237 ]

Answer 6

Another method is using itertools which always comes in useful some way or another:

>>> from itertools import takewhile, ifilter
>>> not_237 = takewhile(lambda L: L != 237, numbers)
>>> is_even = ifilter(lambda L: L % 2 == 0, not_237)
>>> list(is_even)
[402, 984, 360, 408, 980, 544, 390, 984, 592, 236, 942, 386, 462, 418, 344, 236, 566, 978, 328, 162, 758, 918]

So we create a lazy iterator that stops at 237, then take from that even numbers