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I'm finding myself in a predicament where I develop a class then I create an instance of this class. These classes will often perform data and stat operations which take a long time, sometimes 20 minutes.

I'll continue to develop my class and I'll add additional methods to it. Now, how do I update the previous class instance with the new method without reinitializing the older instance of the class, which in this case would take 20 minutes?

martineau
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MadmanLee
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    Does this answer your question? [How to reload the code of a method of class object in Python?](https://stackoverflow.com/questions/4670586/how-to-reload-the-code-of-a-method-of-class-object-in-python) – mkrieger1 Apr 05 '22 at 16:40
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    You need smaller (or pre-cooked) test data to use during development. – chepner Apr 05 '22 at 17:28

1 Answers1

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There's a bit of an XY question here. Your true difficulty seems to be "it takes 20 minutes for the __init__ ctor to complete."

There are several ways out of that predicament.

  1. Spend many minutes computing a new instance, and then arrange to rapidly serialize it to disk. If deserializing takes less than twenty minutes, you win!
  2. Use a level of indirection, so the object that has so many interesting behaviors is lightweight, consuming very little memory. Let it contain pointers to RDBMS id, S3 bucket, url, or filename.
  3. Store the expensive object in a global or equivalent, such as with @lru_cache, and let your updated / reloaded class find it that way. Use importlib.reload().
  4. Similar to serializing, offer a cheap .copy(), and exploit inheritance.
  5. Monkeypatch.

Here is one way to exploit inheritance (copying isn't necessary, but makes it cleaner):

class FooBase:
    ...

# global variable for expensive computation
foo = FooBase()

class FooFeature(FooBase):
    def __init__(self, foo):
        ...
    def some_new_feature(self):
        ...

foo = FooFeature(foo.copy())  # This happens quickly.

Here is one way to monkeypatch:

class Foo:
    ...

foo = Foo()
FooOriginal = Foo

# Now you edit in a brand new Foo feature and reload the class definition:
    def new_feature(self):
        ...

FooOriginal.new_feature = Foo.new_feature

When monkeypatching, note that foo holds a reference to the same class object that FooOriginal holds. Upon reloading, Foo becomes a brand new object, with one additional method. The final monkeypatch assignment makes that method available to FooOriginal, and hence available to foo.

J_H
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