Odd numbers using enumerator

Question

I'm trying to iterate over a list using enumerator in Python, but I want to display only those element that are at odd index of the list. Please note that I do not wish to use the standard step of dividing and checking remainder by 2

Possible duplicate of [Skipping every other element after the first](https://stackoverflow.com/questions/8865878/skipping-every-other-element-after-the-first) — Sayse, Aug 07 '19 at 15:37

keen · Answer 1 · 2019-08-07T15:42:44.290

0

a = range(1,10)
for x, y in enumerate(a):
     if not (x % 2):
         print(y)

edit: downvotes? that's not very nice, y'all

edited Aug 07 '19 at 15:42

answered Aug 07 '19 at 15:39

keen

19
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"I do not wish to use the standard step of dividing and checking remainder by 2" – Sayse Aug 07 '19 at 15:39
he could keep an idx if he wanted and even/odd it, but whatever, i'm pretty sure with the question level, having modulo do the work for him is acceptable – keen Aug 07 '19 at 15:41
@keen can you please elaborate? – Jayashankar Jayan Aug 08 '19 at 16:31

score 0 · Answer 2 · edited Jan 10 '21 at 09:07

0

for i, item in enumerate(list_name[1::2]):...

edited Jan 10 '21 at 09:07

Tomer Shetah

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answered Jan 10 '21 at 07:25

Jayashankar Jayan

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Odd numbers using enumerator

2 Answers2