Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
380 views
in Technique[技术] by (71.8m points)

memoization - Python - anyone have a memoizing decorator that can handle unhashable arguments?

I've been using the following memoizing decorator (from the great book Python Algorithms: Mastering Basic Algorithms in the Python Language ... love it, btw).

def memo(func):
    cache = {}
    @ wraps(func)
    def wrap(*args):
        if args not in cache:
            cache[args] = func(*args)
        return cache[args]
    return wrap

The problem with this decorator is that the dictionary-based cache means that all of my arguments must be hashable.

Does anyone have an implementation (or a tweak to this one) that allows for unhashable arguments (e.g. dictionaries)?

I know that the lack of a hash value means that the question of "is this in the cache?" becomes non-trivial, but I just thought I'd ask.

===EDITED TO GIVE CONTEXT===

I am working on a function that returns a Parnas-style "uses hierarchy" given a dictionary of module: dependencies. Here's the setup:

def uses_hierarchy(requirements):
    """
    uses_hierarchy(requirements)

    Arguments:
    requirements - a dictionary of the form {mod: list of dependencies, }

    Return value:
    A dictionary of the form {level: list of mods, ...}

    Assumptions:
    - No cyclical requirements (e.g. if a requires b, b cannot require a).
    - Any dependency not listed as a mod assumed to be level 0.

    """

    levels = dict([(mod, _level(mod, requirements))
                   for mod in requirements.iterkeys()])
    reversed = dict([(value, []) for value in levels.itervalues()])
    for k, v in levels.iteritems():
        reversed[v].append(k)
    return reversed


def _level(mod, requirements):
    if not requirements.has_key(mod):
        return 0
    dependencies = requirements[mod]
    if not dependencies:
        return 0
    else:
        return max([_level(dependency, requirements)
                    for dependency in dependencies]) + 1

So that:

>>> requirements = {'a': [],
...                 'b': [],
...                 'c': ['a'],
...                 'd': ['a','b'],
...                 'e': ['c','d'],
...                 'f': ['e']
...                 }

>>> uses_hierarchy(requirements)
{0: ['a', 'b'], 1: ['c', 'd'], 2: ['e'], 3: ['f']}

_level is the function I want to memoize to make this setup more scalable. As implemented without memoization, it calculates the level of dependencies multiple times (e.g. 'a' is calculated 8 times I think in the example above).

Thanks,

Mike

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Answer

0 votes
by (71.8m points)

Here is the example in Alex Martelli Python Cookbook that show how to create a memoize decorator using cPickle for function that take mutable argument (original version) :

import cPickle

class MemoizeMutable:
    def __init__(self, fn):
        self.fn = fn
        self.memo = {}
    def __call__(self, *args, **kwds):
        import cPickle
        str = cPickle.dumps(args, 1)+cPickle.dumps(kwds, 1)
        if not self.memo.has_key(str): 
            print "miss"  # DEBUG INFO
            self.memo[str] = self.fn(*args, **kwds)
        else:
            print "hit"  # DEBUG INFO

        return self.memo[str]

Here is a link.

EDIT: Using the code that you have given and this memoize decorator :

_level = MemoizeMutable(_level)

equirements = {'a': [],
               'b': [],
               'c': ['a'],
               'd': ['a','b'],
               'e': ['c','d'],
               'f': ['e']
                 }

print uses_hierarchy(equirements)

i was able to reproduce this:

miss
miss
hit
miss
miss
hit
miss
hit
hit
hit
miss
hit
{0: ['a', 'b'], 1: ['c', 'd'], 2: ['e'], 3: ['f']}

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

2.1m questions

2.1m answers

60 comments

57.0k users

...