"Rogério Brito" <rbr...@ime.usp.br> wrote in message
news:i8lk0n$g3...@speranza.aioe.org...
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
...
v=newlist(1000)
(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...
--
Bartc
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