  lua-l archive

• Subject: Re: Lua Basics: tables as lists?
• From: Rici Lake <lua@...>
• Date: Mon, 22 Aug 2005 22:50:35 -0500

On 22-Aug-05, at 9:52 PM, William Trenker wrote:

It's easy to set up lua lists:
mylist = {"a", "b", "c"}
What I'm wondering is what is the design-intended best practice for
testing for list membership.  And what is the fastest way to do this?

If you don't actually care about the order of the objects (which is a surprisingly common case), then do it like this:

myset = {a = true, b = true, c = true}

-- is an object in the set?
if myset[obj] then ...

-- get all the objects out of the set (in undetermined order):
for k in pairs(myset) do ...

Note that you cannot add objects to the set during the iteration, which is a bit annoying if you're doing a workqueue. However, you can fake it by using three sets.

This assumes that we're storing a digraph as a table of sets, such that
digraph[node_a][node_b] is true if there is a line node_a -> node_b.
Then the following function computes the set of nodes reachable from a given node. It may not be the fastest way of doing it, but it seemed like a simple example:

function reachable(digraph, node)
local empty = {}
local done, work, pending = {}, {}, {}
local function insert(stateset)
if not done[stateset] then
done[stateset], pending[stateset] = true, true
end
end

insert(node)

repeat
pending, work = {}, pending
for node in pairs(work) do
for next_node in pairs(digraph[node] or empty) do
insert(next_node)
end
end
until not next(pending)

return done
end

("not next(pending)" is true if pending is empty, assuming that false is not a possible key. If false were possibly you'd have to say "until nil == next(pending)", which would be more readable but slightly slower.)

Assuming that for-loops are run-time expensive for large lists, is
something like this considered good practice?

For loops are not expensive

myList = {"a", "b", "cde"}
myListString = "|" .. table.concat(myList, "|") .. "|"
a = "cde"
if string.find(myListString, "|"..a.."|") then
print("found!")
else