Populating a list/array by index in Python?

Here's a quick list wrapper that will auto-expand your list with zeros if you attempt to assign a value to a index past it's length.

class defaultlist(list):

   def __setitem__(self, index, value):
      size = len(self)
      if index >= size:
         self.extend(0 for _ in range(size, index + 1))

      list.__setitem__(self, index, value)

Now you can do this:

>>> a = defaultlist([1,2,3])
>>> a[1] = 5
[1,5,3]
>>> a[5] = 10
[1,5,3,0,0,10]

You'll have to pre-fill it with something (e.g. 0 or None) before you can index it:

myList = [None] * 100  # Create list of 100 'None's
myList[12] = 'a'  # etc.

Alternatively, use a dict instead of a list, as Alex Martelli suggested.

Not without populating the other locations in the list with something (like None or an empty string). Trying to insert an element into a list using the code you wrote would result in an IndexError.

There's also mylist.insert, but this code:

myList.insert(12,'a')

would just insert 'a' at the first unoccupied location in the list (which would be 0 using your example).

So, as I said, there has to be something in the list at indexes 0-11 before you can insert something at myList[12].

For a "sparse list" you could use a dict instead:

mylist = {}
mylist[12] = 'a'

etc. If you want an actual list (initialize it with [], not (), of course!-) you need to fill the un-set slots to _some_thing, e.g. None, by a little auxiliary function or by subclassing list.