Removing an element not currently in a list: ValueError?

Question

This is something that's bothered me for a while, and I can't figure out why anyone would ever want the language to act like this:

In [1]: foo = [1, 2, 3]

In [2]: foo.remove(2) ; foo  # okay
Out[2]: [1, 3]

In [3]: foo.remove(4) ; foo  # not okay?
---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)

/home/izkata/<ipython console> in <module>()

ValueError: list.remove(x): x not in list

If the value is already not in the list, then I'd expect a silent success. Goal already achieved. Is there any real reason this was done this way? It forces awkward code that should be much shorter:

for item in items_to_remove:
   try:
      thingamabob.remove(item)
   except ValueError:
      pass

Instead of simply:

for item in items_to_remove:
   thingamabob.remove(item)

---

As an aside, no, I can't just use set(thingamabob).difference(items_to_remove) because I do have to retain both order and duplicates.

Answer 1

From the Python philosophy:

Explicit is better than implicit

The .remove() method has raised an exception from the very first commit; trying to remove a value that is not there gives you explicit feedback instead of implicitly assuming that the value not being there was fine and what you wanted all along.

Answer 2

As an aside: unless your list is extremely large and the list of items to remove is excessively small in comparison, you can create a new list with the items excluded in one readable line:

filtered_list = [x for x in source_list if x not in set(items_to_remove)]

Obviously it preserves order and duplicates as you want. Very probably it is also faster.

Very often not mutating data greatly simplifies the way you think about an algorithm.