python - Setup dictionary lazily -

let's have dictionary in python, defined @ module level (mysettings.py):

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

i values computed when keys accessed:

from mysettings import settings # settings "prepared"  print settings['expensive1'] # value computed. 

is possible? how?

if don't separe arguments callable, don't think it's possible. however, 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 want cache results of expensive_to_compute, if accessed multiple times. 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 @ 0x9b0a62c>, 2), (<function expensive_to_compute @ 0x9b0a62c>, 1)]) >>> settings['expensive1'] 3 >>> settings.values() dict_values([(<function expensive_to_compute @ 0x9b0a62c>, 2), 3]) 

you may want override other dict methods depending of how want use dict.