function test_tableremove(mylist)
    for i=#mylist,1,-1 do
        if mylist[i] == 13 then
            table.remove(mylist, i)
        end
    end
end -- func

function test_mitch(mylist)
    -- NOTE: This is a messy, hardcoded version of my algorithm, for test-purposes.
    -- Go to https://stackoverflow.com/a/53038524/8874388 for the actual, clean version.
    local j, n = 1, #mylist;

    for i=1,n do
        local value = mylist[i]
        if value ~= 13 then -- Keep everything except "13"
            -- Move i's kept value to j's position, if it's not already there.
            if (i ~= j) then
                mylist[j] = mylist[i];
                mylist[i] = nil;
            end
            j = j + 1; -- Increment position of where we'll place the next kept value.
        else -- Delete "13"
            mylist[i] = nil;
        end
    end
end

function build_tables()
    local tables = {}
    for i=1, 10 do
      tables[i] = {}
      for j=1, 2000000 do -- 2 million items
        tables[i][j] = (j % 5 == 0 and 13 or 100) -- put a "13" (deleted) in every 5th item, otherwise "100" (kept)
      end
    end

    return tables
end

function time_func(func, name)
    local tables = build_tables()
    time0 = os.clock()
    for i=1, #tables do
        func(tables[i])
    end
    time1 = os.clock()
    print(string.format("[%10s] elapsed time: %.3f\n", name, time1 - time0))
end

time_func(test_mitch, "mitch")
time_func(test_tableremove, "table.remove")
time_func(test_mitch, "mitch")
time_func(test_tableremove, "table.remove")