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 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 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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