Trying to understand Python memoization code snippet

Question

In a recent Hacker Newsletter issue, this very useful article about decorators in Python was linked. I like the article and I think I understand most of the decorator examples. However, in the non-decorator memoization example, I’m very confused by the code:

def memoize(fn):
    stored_results = {}

    def memoized(*args):
        try:
            # try to get the cached result
            return stored_results[args]
        except KeyError:
            # nothing was cached for those args. let's fix that.
            result = stored_results[args] = fn(*args)
            return result

    return memoized

I’m confused about how this function would create a persisting dictionary stored_results that gets appended to. After re-reading it, copy/pasting it into my editor and playing with it, and looking online for help, I still don’t understand what the syntax stored_results[args] = fn(*args) is really doing.

(1) The article suggests that the above code will return the function, but that now it will search a dictionary first before executing on novel arguments. How does this happen? Why isn’t stored_results just local to memoize? Why doesn’t it get destroyed when memoized is returned?

(2) Links to other questions or web resources that explain the argument passing here with *args would be helpful too. If *args is a list of arguments, why can we use the syntax stored_results[args], when normally you get a non-hashable error when trying to index a dictionary on a list?

Thanks for any clarifying thoughts.

Asked By: ely

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Answer 1

If *args is a list of arguments, why can we use the syntax stored_results[args], when normally you get a non-hashable error when trying to index a dictionary on a list?

Because it’s not a list, it’s a tuple. Lists are mutable, so you can’t define a meaningful hash function on them. Tuples however are immutable data structures.

Why isn’t stored_results just local to memoize? Why doesn’t it get destroyed when memoized is returned?

Because memoize and memoized share the same name context (closure). The closure persists because memoized holds a reference to it and you return it and assign it to a global name (this is the effect of the decorator statement). With the closure, all the captured values persist as well.

Answered By: Niklas B.

Trying to understand Python memoization code snippet

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