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Using the * Operator The * operator can also be used to repeat elements of a list. When we multiply a list with any number using the * operator, it repeats the elements of the given list.
To initialize a list in Python assign one with square brackets, initialize with the list() function, create an empty list with multiplication, or use a list comprehension. The most common way to declare a list in Python is to use square brackets.
As mentioned above, in Python, you can easily add and remove elements from a list, so in most cases, it is not necessary to initialize the list in advance.
The first thing that comes to mind for me is:
verts = [None]*1000
But do you really need to preinitialize it?
Not quite sure why everyone is giving you a hard time for wanting to do this - there are several scenarios where you'd want a fixed size initialised list. And you've correctly deduced that arrays are sensible in these cases.
import array
verts=array.array('i',(0,)*1000)
For the non-pythonistas, the (0,)*1000 term is creating a tuple containing 1000 zeros. The comma forces python to recognise (0) as a tuple, otherwise it would be evaluated as 0.
I've used a tuple instead of a list because they are generally have lower overhead.
One obvious and probably not efficient way is
verts = [0 for x in range(1000)]
Note that this can be extended to 2-dimension easily. For example, to get a 10x100 "array" you can do
verts = [[0 for x in range(100)] for y in range(10)]
Wanting to initalize an array of fixed size is a perfectly acceptable thing to do in any programming language; it isn't like the programmer wants to put a break statement in a while(true) loop. Believe me, especially if the elements are just going to be overwritten and not merely added/subtracted, like is the case of many dynamic programming algorithms, you don't want to mess around with append statements and checking if the element hasn't been initialized yet on the fly (that's a lot of code gents).
object = [0 for x in range(1000)]
This will work for what the programmer is trying to achieve.
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