Re: filter an array in place

["Robert G. Jakabosky" <bobby@...>]

On Thursday 13, Axel Kittenberger wrote:
> You must not change a table while iterating through it with
> for/ipairs/pairs. While it might work with some reversal its still bad
> style can possibly break.

How can using a for loop counting down break?  Removing a node only effects 
the position of the nodes that where already processed.  The original 
question was for a "more elegant way" to solve the problem.

> This will work instead:
>
> local tab = {"a", "a", "c", "a", "d", "a" }
>
> local len = #tab
> local k = 1
> while tab[k] do
>     if tab[k] == "a" then
>         table.remove(tab, k)
>     else
>         k = k + 1
>     end
> end
>
> for k, v in ipairs(tab) do print(k, v) end
> print()
> for k, v in pairs(tab) do print(k, v) end
>
>
>
> On Fri, Jan 14, 2011 at 5:26 AM, Robert G. Jakabosky
>
> <bobby@sharedrealm.com> wrote:
> > On Thursday 13, Emmanuel Oga wrote:
> >> Hello, I'm looking for the best proper way of doing this:
> >>
> >> local tab = {"a", "a", "c", "a", "d", "a" }
> >>
> >> for index, val in ipairs(tab) do
> >>   if val == "a" then -- could be any other complex condition
> >>     table.remove(tab, index)
> >>   end
> >> end
> >>
> >> -- WRONG: tab = { "a", "c", "d" }, should be { "c", "d" }
> >>
> >> So removing elements from a container _while you are iterating it_ is
> >> wrong in most languages. I'm looking for the nicest and proper way of
> >> doing it.
> >>
> >> Two methods I came up with:
> >>
> >> 1) popping from the array and pushing to a new one if the condition is
> >> met. 2) Iterating first, storing indexes to remove, then removing one by
> >> one (compensating index offset as a result of each removed elements)
> >>
> >> http://pastie.org/1458808
> >>
> >> Is there any other more elegant way?
> >
> > process the list in reverse.
> >
> > for index=#tab,1,-1 do
> >  local var = tab[index]
> >   if val == "a" then -- could be any other complex condition
> >     table.remove(tab, index)
> >   end
> > end
> >
> >
> > --
> > Robert G. Jakabosky



-- 
Robert G. Jakabosky