How to begin from start once index out of range - Python

Question

I have a list of items:

items = [a, b, c, d]

I need it to be so that if I say items[4] which isn't in the list, the returned element should be the first element only, which is a, like a loop, instead of a IndexError: index out of range error.

Does this answer your question? [Wrapping around on a list when list index is out of range](https://stackoverflow.com/questions/22122623/wrapping-around-on-a-list-when-list-index-is-out-of-range) — mkrieger1, Dec 19 '20 at 09:38

quamrana · Accepted Answer · 2020-12-19T10:16:36.103

2

You can employ operator %:

def getItem(items, index):
    return items[index % len(items)]

items = [‘a’, ‘b’, ‘c’, ‘d’]
print(getItem(items, 42))

Output:

edited Dec 19 '20 at 10:16

answered Dec 19 '20 at 09:31

quamrana

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score 0 · Answer 2 · answered Dec 19 '20 at 09:47

0

To avoid the error message, you can handle it as an exception for the indexes not in array's range. i.e. if the index entered is n then

try:
    print(items[n])
except:
    print(items[n % len(items)])

I hope this helped..!

answered Dec 19 '20 at 09:47

DNShah

46
3

How to begin from start once index out of range - Python

2 Answers2