Sparse assignment list in python

Question

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 library.

Alex Martelli · Accepted Answer · 2009-12-07 04:45:42Z

26

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.

answered Dec 7, 2009 at 4:45

Alex Martelli

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4 Comments

Stefano Borini Over a year ago

However, this means that it does not exist in the standard library... may I include your code in the wavemol (BSD) library ?

Alex Martelli Over a year ago

@Stefano, sure, see meta.stackexchange.com/questions/13976/… : "user contributed content licensed under cc-wiki with attribution required" per Jeff Attwood's answer (I believe the BSD license is compatible w/that, but I'll be happy to relicense it to you otherwise as needed!-).

Jeff Learman Over a year ago

Do Python lists automatically optimize space for this?

Jeff Learman Over a year ago

A little testing shows they don't, so while this meets the OP's stated requirements, it doesn't meet the usual expectations of a sparse array.

wjandrea · Accepted Answer · 2020-12-16 16:29:48Z

7

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'

edited Dec 16, 2020 at 16:29

wjandrea

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answered Jan 7, 2013 at 8:02

htmldrum

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1 Comment

Arthur Khazbulatov Over a year ago

If you need to iterate over the sparse array items and the order is significant, a dictionary-based solution would be inefficient

Erik · Accepted Answer · 2022-04-30 14:17:18Z

0

The sparse_list package provides the behaviour OP asks for.

answered Apr 30, 2022 at 14:17

Erik

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