Unregister for singledispatch?

Question

Is there way to "unregister" a registered function for a generic ?

For example:

from functools import singledispatch

@singledispatch
def foo(x):
    return 'default function'

foo.register(int, lambda x: 'function for int')

# later I would like to revert this.

foo.unregister(int) # does not exist - this is the functionality I am after

Answer 1

singledispatch is meant to be append only; you cannot really unregister anything.

But as with all things Python, the implementation can be forced to unregister. The following function will add a unregister() method to a singledispatch function:

def add_unregister(func):
    # build a dictionary mapping names to closure cells
    closure = dict(zip(func.register.__code__.co_freevars, 
                       func.register.__closure__))
    registry = closure['registry'].cell_contents
    dispatch_cache = closure['dispatch_cache'].cell_contents
    def unregister(cls):
        del registry[cls]
        dispatch_cache.clear()
    func.unregister = unregister
    return func

This reaches into the closure of the singledispatch.register() function to access the actual registry dictionary so we can remove an existing class that was registered. I also clear the dispatch_cache weak reference dictionary to prevent it from stepping in.

You can use this as a decorator:

@add_unregister
@singledispatch
def foo(x):
    return 'default function'

Demo:

>>> @add_unregister
... @singledispatch
... def foo(x):
...     return 'default function'
... 
>>> foo.register(int, lambda x: 'function for int')
<function <lambda> at 0x10bed6400>
>>> foo.registry
mappingproxy({<class 'object'>: <function foo at 0x10bed6510>, <class 'int'>: <function <lambda> at 0x10bed6400>})
>>> foo(1)
'function for int'
>>> foo.unregister(int)
>>> foo.registry
mappingproxy({<class 'object'>: <function foo at 0x10bed6510>})
>>> foo(1)
'default function'