Sparse assignment list in python

I need a list with the following behavior

>>> l = SparseList()
>>> l
[]
>>> l[2] = "hello"
>>> l
[ None, None, "hello"]
>>> l[5]
None
>>> l[4] = 22
>>> l
[ None, None, "hello", None, 22]
>>> len(l)
5
>>> for i in l: print i
None
None
"hello"
None
22

Although it can "emulated" via a dictionary, it's not exactly the same. numpy array can behave this way, but I don't want to import the whole numpy for something like this. Before coding it myself, I ask if something similar exists in the standard lib开发者_运维技巧rary.

---

Here's minimal code to pass your given examples (with indispensable adjustments: you expect weird spacing and quoting, 'None' to be printed out at the prompt without a print statement, etc etc):

class SparseList(list):
  def __setitem__(self, index, value):
    missing = index - len(self) + 1
    if missing > 0:
      self.extend([None] * missing)
    list.__setitem__(self, index, value)
  def __getitem__(self, index):
    try: return list.__getitem__(self, index)
    except IndexError: return None

__test__ = dict(allem='''
>>> l = SparseList()
>>> l
[]
>>> l[2] = "hello"
>>> l
[None, None, 'hello']
>>> print l[5]
None
>>> l[4] = 22
>>> l
[None, None, 'hello', None, 22]
>>> len(l)
5
>>> for i in l: print i
None
None
hello
None
22
''')
import doctest
doctest.testmod(verbose=1)

I imagine you'll want more (to support negative indices, slicing, and whatever else), but this is all your examples are implicitly specifying.

---

Dictionaries can be used as sparse lists. Whilst they will not provide the characteristics you are after (as you are not actually after a sparse list, all the list elements are complete references to None in a dynamically-sized Array), they act as a textbook sparse array.

sparse_vars = [(0,"Hi"), (10000,"Bye"), (20000,"Try")]
sparse_list = {}

for var in sparse_vars:
  sparse_list[var[0]] = var[1]

>>> print sparse_list
{0: 'Hi', 10000: 'Bye', 20000: 'Try'}
>>> print sparse_list[20000]
'Try'

---

The sparse_list package provides the behaviour OP asks for.