Thanks very much. Is there a way to do that without the list comprehension? I'm building this as part of a class, and I will not have talked about list comprehensions up until that point.
Thanks, Irv > On May 20, 2018, at 2:35 PM, Daniel Foerster <pydsig...@gmail.com> wrote: > > I think what you're looking for is this: > > my_list = [x for x in my_list if not need_to_delete(x)] > > On Sun, May 20, 2018, 16:28 Irv Kalb <i...@furrypants.com > <mailto:i...@furrypants.com>> wrote: > I am building a game where I am keeping track of many objects in a list > (let's just call it "myList". In every "frame" of my game I need to see if > any objects need to be removed from myList. > > I know that this code: > > for item in myList: > if needsToBeDeleted(item): > myList.remove(item) > > does not work correctly because it affects the order of list while I am > iterating over it. (Specifically, if there are two elements in a row that > should be removed, the first one is removed, but the second one is skipped > over.) > > I have looked at many different solutions, and have found a number of > different approaches, but none them seem elegant. Here are ones that work: > > 1) Build a list of indices in reverse order ( [n - 1, n -2, n - 3, ... 0] ) > Iterate over this list > if the element at that index needs to be deleted, > pop the element at this index > > (untested, but should work) > > indexList = list(range(0, n - 1)) > indexList.reverse() > > for index in indexList: > if needsToBeDeleted(myList[index]): > myList.pop(index) > > I believe this will work, but it looks very clunky. > > > 2) Create a tempList that starts off empty. > Iterate through my original list. > If an element does NOT need to be deleted, append it to the tempList. > > Reassign the myList to the tempList > Set the tempList to None (or delete it) > > tempList = [] > for item in myList: > if not needsToBeDeleted(item): > tempList.append(item) > myList = tempList > tempList = None > > Seems OK, but again, seems like kind of a kludge. > > 3) Iterate through a copy of the list: > > for item in myList[:]: > if needsToBeDeleted(item): > myList.remove(item) > > This does work, but I would guess it might be slow for long lists > (since it first has to duplicate a list, then when it wants to delete an > element, 'remove' must search through the list to find the matching element > before deleting it). Also, it would be tricky to explain to students, > because I am iterating through a copy of a list, but deleting elements from > the original list. > > > Does anyone have an easier, more clear, reusable approach? > > If not, any thoughts on which of these three approach would be most efficient > and clear? I don't have a favorite yet. > > Thanks, > > Irv > >
