Setup dictionary lazily

Question

Let's say I have this dictionary in python, defined at the module level (mysettings.py):

settings = {
    'expensive1' : expensive_to_compute(1),
    'expensive2' : expensive_to_compute(2),
    ...
}

I would like those values to be computed when the keys are accessed:

from mysettings import settings # settings is only "prepared"

print settings['expensive1'] # Now the value is really computed.

Is this possible? How?

Answer 1

If you don't separe the arguments from the callable, I don't think it's possible. However, this should work:

class MySettingsDict(dict):

    def __getitem__(self, item):
        function, arg = dict.__getitem__(self, item)
        return function(arg)


def expensive_to_compute(arg):
    return arg * 3

And now:

>>> settings = MySettingsDict({
'expensive1': (expensive_to_compute, 1),
'expensive2': (expensive_to_compute, 2),
})
>>> settings['expensive1']
3
>>> settings['expensive2']
6

Edit:

You may also want to cache the results of expensive_to_compute, if they are to be accessed multiple times. Something like this

class MySettingsDict(dict):

    def __getitem__(self, item):
        value = dict.__getitem__(self, item)
        if not isinstance(value, int):
            function, arg = value
            value = function(arg)
            dict.__setitem__(self, item, value)
        return value

And now:

>>> settings.values()
dict_values([(<function expensive_to_compute at 0x9b0a62c>, 2),
(<function expensive_to_compute at 0x9b0a62c>, 1)])
>>> settings['expensive1']
3
>>> settings.values()
dict_values([(<function expensive_to_compute at 0x9b0a62c>, 2), 3])

You may also want to override other dict methods depending of how you want to use the dict.