GeezerVison

How likely is it that the object is in user_ns? If it's only occasionally, can wrap this in:

if graphic in ip.user_ns.values():
    ...

I guess it's possible that the graphic is present more than once, under a different name?

Alt solution that might be faster because it avoids copying the keys and looking up the values:

            keys_to_delete = []
            for key, ob in ip.user_ns.items():
                if ob is graphic:
                    keys_to_delete.append(key)
            for key in keys_to_delete:
                    ip.del_var(key)

How likely is it that the object is in user_ns? If it's only occasionally, can wrap this in:

if graphic in ip.user_ns.values():
    ...

in operators don't work if any item of the iterable is a numpy array, for example:

a = np.random.rand(10)
l = [a, "bah", 123]

# will raise a ValueError
123 in l

I've resorted to this, couldn't think of a more performance solution. Time taken scales with the number of items in the namespace.

if IS_IPYTHON:
    # remove any references that ipython might have made
    ip = get_ipython()

    # check both namespaces
    for namespace in [ip.user_ns, ip.user_ns_hidden]:
        # find the reference
        for ref, obj in namespace.items():
            if graphic is obj:
                # we found the reference, remove from ipython
                ip.del_var(ref)
                break

No problems with deleting items from an object being iterated over? I guess it depends on IPython internals whether that'd be a problem.