while loop - IndexError: list index out of range

Question

Maybe it's too simple and I just didn't see my mistake.

while list_a[y] in list_a != list_a[-1]:
    print(y);y=y+1

returns IndexError: list index out of range

list_a looks like:

['00001', '00001', '00002', '00009', '0000G', '0000K', '0000K', '0000U', '0000U', '00013', '0001B', '0001D', '0001D', '0001L', '0001L', '0001N', '0001Q', '0001Q', '0001R', '0001U']

and my aim in the end is to delete some items from the list while iterating (that's why I want to use a while loop instead of for y in range(len(list_a))).

What do you think this condition does? It tests whether list_a[y] is in list_a (either true or indexerror) and then compares that (true) to list_a[-1]. Probably not what you wanted. Have you tried just using while y < len(list_a)? — tobias_k
– tobias_k, Commented Jun 19, 2014 at 7:51
in the end the loop does <something> until the last entry of the list everything correctly. Then it compares again and the error occurs — user3753064
– user3753064, Commented Jun 19, 2014 at 7:55
(Minor correction: in and != have the same precedence, so x in y != z actually evaluates like (x in y) and (y != z), similar to x < y < z.) — tobias_k
– tobias_k, Commented Jun 19, 2014 at 8:12

jonrsharpe · Accepted Answer · 2014-06-19 08:14:25Z

2

Think what you were trying for was:

while list_a[y] != list_a[-1]:
    ...

i.e. "while we're looking at the item that isn't equal to the last in the list". However, there will still be issues; what if items appear elsewhere in the list that are equal to the last item?

The general way to do this is to use a list comprehension to build a new list from the appropriate items in the old list:

list_b = [item for item in list_a if some_test(item)]

edited Jun 19, 2014 at 8:14

answered Jun 19, 2014 at 7:59

jonrsharpe

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