How to remove all objects in a list from a for loop with python?

Question

I feel like this should be a pretty simple process but for some reason I cannot get it to work.

Goal: Loop through a list of objects and if they match a condition remove it.

Issue: This works until it needs to remove all objects from the list. It will not remove the last item.

Example (clearly simplified by same results):

def some_function(val): return True on_shift2 = ['user1', 'user2', 'user3'] print("MY ON SHIFT2", on_shift2) for idx, val in enumerate(on_shift2): print(f"Index: {idx} & Value: {val}") if len(on_shift2) == 1: del (on_shift2) # remove the whole list else: check_user = some_function(val) # function to check if user is in list and if so return true if check_user: on_shift2.remove(val) # Also tried the following # del on_shift[idx] # Same result # on_shift.pop(idx) # Same result print("MY ON SHIFT2", on_shift2)

Results:

MY ON SHIFT2 ['user1', 'user2', 'user3'] Index: 0 & Value: user1 Index: 1 & Value: user3 MY ON SHIFT2 ['user2']

Now the really odd part here is no matter what is in the list, it always seems to keem the middle item (i.e. user2). If i changed those values to chicken, horse, dog - the remaining item would be horse. Not sure if that's important but thought i would bring it up.

Also, no matter the which method i choose to remove the the items from the list (pop, remove or del), the 3rd (or last) item never gets removed if they all return true from "some_function()".

Also when looping through the if condition on the len(on_shift) never does equal 1, it always stops at 2. Never makes it to the 3rd (or last) object if the all return true from "some_fuction()", so my if statement to delete the list never happens.

If i print the val for the loop without removing anything from the list, all 3 names show up. I have to be doing something silly here.

Thoughts?

you should not remove from a list while iterating, if you must do so , then for idx, val in enumerate(on_shift2.copy()) copy and then delete, your if condition if len(on_shift2) == 1 is not needed anymore — python_user
– python_user, Commented Jun 8, 2021 at 2:37
I'm getting a different error: NameError: name 'on_shift2' is not defined — David Kaftan
– David Kaftan, Commented Jun 8, 2021 at 2:39
try without if check_user: line and see what happens. I guess the problem is in this line it returns false sometimes and all items do not get deleted. — Sudipto
– Sudipto, Commented Jun 8, 2021 at 2:41
@phython_user thank you for the quick reply. This is what i needed. If you want to set this as an answer ill mark it as correct. @David I get that error when i use .copy() and leave the code the same as the list gets removed in the if len(on_shift2) clause. — jAC
– jAC, Commented Jun 8, 2021 at 2:43
@Sudipto at times what you are saying is true and that part was working fine, i just couldn't figure out how to get it to remove all them when it was 100% true. — jAC
– jAC, Commented Jun 8, 2021 at 2:44

python_user · Accepted Answer · 2021-06-08 03:04:27Z

You should not modify a list while iterating over it, but if you must do so, then you need to a make a copy of the list and iterate over the copy, but delete the elements from the original list.

def some_function(val): return True on_shift2 = ['user1', 'user2', 'user3'] print("MY ON SHIFT2", on_shift2) for idx, val in enumerate(on_shift2.copy()): # copy here print(f"Index: {idx} & Value: {val}") check_user = some_function(val) # function to check if user is in list and if so return true if check_user: on_shift2.remove(val) # delete in original print("MY ON SHIFT2", on_shift2)

Output

MY ON SHIFT2 ['user1', 'user2', 'user3'] Index: 0 & Value: user1 Index: 1 & Value: user2 Index: 2 & Value: user3 MY ON SHIFT2 []

This is actually slower, a better alternative would be to add only those you need to a different list.

on_shift2_new = [i for i in on_shift2 if not some_function(i)]

great alternative solution as well! I will test with both to see if performance is an issue and if so i will move the "false" items to a new list rather than do the copy.

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