Re: filter an array in place

[Steve Litt <slitt@...>]

On Thursday 13 January 2011 23:12:40 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?

I'd have a variable called goes_here initialized to 1. Each iteration of your 
loop increments goes_here UNLESS tab[goes_here] meets the conditon. On each 
iteration, copy tab[index] to tab[goes_here]. At the end goes_here indicates 
either the last valid entry or the first invalid one.

Obviously I haven't thought all of this out, but you know what I mean, and it 
will do it without deleting or nilling a key, and it should be pretty darn 
fast because you do it in place and loop only once.

SteveT

Steve Litt
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