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

--

*------------------------------[image: http://i153.photobucket.com/albums/s202/animatics/probiner-sig.gif%5DPedro Alpiarça dos Santos Animator 3DModeler Illustrator >> http://probiner.x10.mx/ http://probiner.x10.mx/*

-- Only when you actually get to the state where there is neither delusion nor enlightenment are you finally comfortable...Foyan