Hi David

Thanks for the idea. Just some considerations:

- Del is a list with indices and not the items themselves, so if you replace the code with that line it won't output the same result.
- del deletes all items with an equal value so if I have L=[2,2,2,1,2] and have Del=[2], I end up with L=[1], which useful, but not my goal here and the reason I went to pop() instead, so I can address indices. Same remark for in.
- I wouldn't mind if in and del approach would work if you instead of checking the value would be checking if a list had such indices and then deleting all at the same time if so.

Again the same code with comments. http://repl.it/f6E/3

Thanks

On Thu, Mar 26, 2015 at 2:04 AM, David Blaschke <dwblas@gmail.com> wrote:

ist comprehension is straight forward IMHO, and possibly faster than
pop depending on whether the items are removed one at a time or just
marked for deletion later
new_list=[item for item in L if item not in Del]

On 3/25/15, pedro santos <probiner@gmail.com> wrote:
> Is this best way to remove multiple items from a list?
> Because if so I think passing a list of integers through pop would be
> cleaner.
>
> Code here: http://repl.it/f6E
>
> Cheers
>
>
>
>
> --
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