creating a defaultlist in python

Question

I\'m trying to create a list equivalent for the very useful collections.defaultdict. The following design works nicely:

class defaultlist(list):
    def __in         


        

Answer 1

On the example you give, you first try to retrieve a non-existing value on the list, as you do dl[2]['a'], Python first retrieve the third (index 2) element on the list, then proceed to get the element named 'a' on that object - therefore you have to implement your automatic extending behavior to the __getitem__ method as well, like this:

class defaultlist(list):
    def __init__(self, fx):
        self._fx = fx
    def _fill(self, index):
        while len(self) <= index:
            self.append(self._fx())
    def __setitem__(self, index, value):
        self._fill(index)
        list.__setitem__(self, index, value)
    def __getitem__(self, index):
        self._fill(index)
        return list.__getitem__(self, index)

Answer 2

There is a python package available:

$ pip install defaultlist

Added indicies are filled with None by default.

>>> from defaultlist import defaultlist
>>> l = defaultlist()
>>> l
[]
>>> l[2] = "C"
>>> l
[None, None, 'C']
>>> l[4]
>>> l
[None, None, 'C', None, None]

Slices and negative indicies are supported likewise

>>> l[1:4]
[None, 'C', None]
>>> l[-3]
'C'

Simple factory functions can be created via lambda.

>>> l = defaultlist(lambda: 'empty')
>>> l[2] = "C"
>>> l[4]
'empty'
>>> l
['empty', 'empty', 'C', 'empty', 'empty']

It is also possible to implement advanced factory functions:

>>> def inc():
...     inc.counter += 1
...     return inc.counter
>>> inc.counter = -1
>>> l = defaultlist(inc)
>>> l[2] = "C"
>>> l
[0, 1, 'C']
>>> l[4]
4
>>> l
[0, 1, 'C', 3, 4]

See the Documentation for any further details.