This question has many answers already, but all seem inferior to my favorite approach to this problem, which iterates through and tests each item just once, and uses the speed of list-comprehension to build one of the two output lists, so that it only has to use the comparatively slow append to build one of the output lists:
bad = []good = [x for x in mylist if x in goodvals or bad.append(x)]In my answer to a similar question, I explain how this approach works (a combination of Python's greedy evaluation of or refraining from executing the append for "good" items, and append returning a false-like value which leaves the if condition false for "bad" items), and I show timeit results indicating that this approach outcompetes alternatives like those suggested here, especially in cases where the majority of items will go into the list built by list-comprehension (in this case, the good list).
