Many newbie questions regarding python

BartC bc at
Fri Oct 8 14:24:32 CEST 2010

"Rogério Brito" <rbrito at> wrote in message 
news:i8lk0n$g38$1 at

> My first try to write it in Python was something like this:
> v = []
> for i in range(20):
>    v[i] = 0
> Unfortunately, this doesn't work, as I get an index out of bounds when 
> trying to
> index the v list.

Python can't grow a list by assigning to out-of-bound elements (although, 
being the language it is, there is probably a way of achieving that by 
redefining how [] works...)

> What is the Pythonic way of writing code like this? So far, I have found 
> many

>    v = [0 for i in range(20)]
>    v = [0] * 20
>    v = []
>    for i in range(20): v.append(0)
> What should I prefer? Any other alternative?

v=[0,0,0,0,0, 0,0,0,0,0, 0,0,0,0,0, 0,0,0,0,0]

will also work. But none of these appeal too much. I would probably do:

def newlist(length,init=0):
        return [init]*length

(with the proviso that someone mentioned when the init value is complex: you 
might not get unique copies of each).

> If possible, I would like to simply declare the list and fill it latter in 
> my
> program, as lazily as possible (this happens notoriously when one is using 
> a
> technique of programming called dynamic programming where initializing all
> positions of a table may take too much time in comparison to the filling 
> of the
> array).

A sparse array? Even if an array could be constructed by assigning to 
arbitrary elements, the gaps created would still need filling in with None 
or Unassigned.

> 2 - If I declare a class with some member variables, is is strictly 
> necessary

This is where I bail out...


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