Python: Bind an Unbound Method?

All functions are also descriptors, so you can bind them by calling their __get__ method:

bound_handler = handler.__get__(self, MyWidget)

Here's R. Hettinger's excellent guide to descriptors.

As a self-contained example pulled from Keith's comment:

def bind(instance, func, as_name=None):    """    Bind the function *func* to *instance*, with either provided name *as_name*    or the existing name of *func*. The provided *func* should accept the     instance as the first argument, i.e. "self".    """    if as_name is None:        as_name = func.__name__    bound_method = func.__get__(instance, instance.__class__)    setattr(instance, as_name, bound_method)    return bound_methodclass Thing:    def __init__(self, val):        self.val = valsomething = Thing(21)def double(self):    return 2 * self.valbind(something, double)something.double()  # returns 42

This can be done cleanly with types.MethodType. Example:

import typesdef f(self): print selfclass C(object): passmeth = types.MethodType(f, C(), C) # Bind f to an instance of Cprint meth # prints <bound method C.f of <__main__.C object at 0x01255E90>>

Creating a closure with self in it will not technically bind the function, but it is an alternative way of solving the same (or very similar) underlying problem. Here's a trivial example:

self.method = (lambda self: lambda args: self.do(args))(self)