Python: Bind an Unbound Method?

Question

In Python, is there a way to bind an unbound method without calling it?

I am writing a wxPython program, and for a certain class I decided it'd be nice to group the data of all of my buttons together as a class-level list of tuples, like so:

class MyWidget(wx.Window):
    buttons = [("OK", OnOK),
               ("Cancel", OnCancel)]

    # ...

    def Setup(self):
        for text, handler in MyWidget.buttons:

            # This following line is the problem line.
            b = wx.Button(parent, label=text).Bind(wx.EVT_BUTTON, handler)

The problem is, since all of the values of handler are unbound methods, my program explodes in a spectacular blaze and I weep.

I was looking around online for a solution to what seems like should be a relatively straightforward, solvable problem. Unfortunately I couldn't find anything. Right now, I'm using functools.partial to work around this, but does anyone know if there's a clean-feeling, healthy, Pythonic way to bind an unbound method to an instance and continue passing it around without calling it?

Answer 1

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_method

class Thing:
    def __init__(self, val):
        self.val = val

something = Thing(21)

def double(self):
    return 2 * self.val

bind(something, double)
something.double()  # returns 42