Autogrowing list in Python?

Question

I need a list-like object that will "autogrow" whenever a slot number greater or equal to its length is accessed, filling up all the newly created slots with some pre-specified default value. E.g.:

# hypothetical DefaultList class
x = DefaultList(list('abc'), default='*')
x[6] = 'g'
print x[2], x[4], x[6], x[8]  # should print 'c * g *'

Thanks!

PS. I know it is not hard to implement a class like this, but I avoid wheel-reinvention as much as possible, especially if a particularly efficient/well-designed wheel already exists.

PS2. A dict (or a collections.defaultdict) is not an acceptable implementation of the desired data structure. For why, see here: http://groups.google.com/group/comp.lang.python/msg/bcf360dfe8e868d1?hl=en

Answer 1

class DefaultList(list):
    def __init__(self,*args,**kwargs):
        list.__init__(self,*args)
        self.default=kwargs.get('default',None)
    def __getitem__(self,key):
        # retrieving an item does not expand the list
        if isinstance(key,slice):
            return [self[elt] for elt in range(key.start,key.stop,key.step)]
        else:
            try:
                return list.__getitem__(self,key)
            except IndexError:
                return self.default
    def __setitem__(self,key,value):
        # setting an item may expand the list
        try:
            list.__setitem__(self,key,value)
        except IndexError:
            self.extend([self.default]*(key-len(self)))
            self.append(value)

x = DefaultList(list('abc'), default='*')
print(x)
# ['a', 'b', 'c']
x[6] = 'g'
print(x)
# ['a', 'b', 'c', '*', '*', '*', 'g']
print x[2], x[4], x[6], x[8]  # should print 'c * g *'
# c * g *
print(x[2:9:2])
# ['c', '*', 'g', '*']

Answer 2

I would use a sparse data structure (1xn matrix).