Reorder table element (SOLVED)

Questions
#1

I have

global_.game_data.turn.step = {
[22] = 0,
[3] = 0,
[2] = 1,
[4] = 0,
}

i want

global_.game_data.turn.step = {
[2] = 1,
[4] = 0,
[22] = 0,
[3] = 0,
}

or I have

global_.game_data.turn.step = {
[22] = 0,
[3] = 0,
[2] = 0,
[4] = 1,
}

i want

global_.game_data.turn.step = {
[4] = 1,
[22] = 0,
[3] = 0,
[2] = 0,
}

or I have

global_.game_data.turn.step = {
[4] = 0,
[22] = 1,
[3] = 0,
[2] = 0,
}

i want

global_.game_data.turn.step = {
[22] = 1,
[3] = 0,
[2] = 0,
[4] = 0,
}

just move element with no zero to top and circle reorder with zero
or this imposible?

#2

This is all Lua related.

Printing (i.e. iterating on the keys of) a Lua table isn’t guarantueed to output a sorted list of keys, as it’s a hash table, and their storage details are internal to Lua.

You could store the items as pairs instead, then it should be easier to sort them:

x = { {4,0}, {22,1}, {3,0}, {2,0} }
table.sort(x, function (a,b) return a[2] > b[2]; end)
for k,v in ipairs(x) do print(k, v[1], v[2]) end
1 Like
#3

Many thanks ))

yes element with 1 value go to top corectly

1	22	1
2	3	0
3	4	0
4	2	0

how about other element

i need

1	22	1
2	3	0
3	2	0
4	4	0
#4

very strange languge ((

I need this to set the Z coordinate depending on the player’s turn order

#5

Not sure how to sort that? It feels like something is missing for that to happen?
I.e. how would you know that 3 should be sorted before 2 and 4?

However, once you do know that, you can simply modify your sort function to sort on each value at a time. E.g.:

table.sort(x, function (a,b)
                    if a[2] == b[2] then
                        return a[1] < b[1];
                    else
                        return a[2] > b[2];
                    end;
                end)

very strange languge

In which sense?
Perhaps that would help if you could specify what is missing?

#6

in order after the element with one

i have this table and i need use in loop

x = { {4,1}, {22,0}, {3,0}, {2,0} } -- do nothing

x = { {4,0}, {22,1}, {3,0}, {2,0} } -- have
x = { {22,1}, {3,0}, {2,0} , {4,0}} -- need - just top three element  {22,1}, {3,0}, {2,0} and set {4,0} to last

x = { {4,0}, {22,0}, {3,1}, {2,0} } -- have
x = { {3,1}, {2,0}, {4,0} , {22,0}} -- need - just top two element  {3,1}, {2,0}  and set {4,0}, {22,0} to last

x = { {4,0}, {22,0}, {3,0}, {2,1} } -- have
x = { {2,1}, {4,0}, {22,0} , {3,0}} -- need - just top one element  {2,1} and set {4,0}, {22,0}, {3,0} to last

and so on in a circle to make a shift

#7

you need something like this

local function sort(x)
	local index = 0
	for i = 1, #x do
		if x[i][2] == 1 then
			index = i
			break
		end
	end
	if index > 1 then
		local ret = {}
		for i = index, #x do
			table.insert(ret, x[i])
		end
		for i = 1, index - 1 do
			table.insert(ret, x[i])
		end
		return ret
	else
		return x
	end
end

function init(self)
	local x = { {4,0}, {22,1}, {3,0}, {2,0} }
	pprint(sort(x))
end
2 Likes
#8

Yes )) it work

thank you very much

#9

RIght, so I think I understand now.
So to describe the algorithm, it’d be something like:

  • Find index of new “head” of the list, store in count
  • Rotate the list left count steps

What @Chung_Xa outlined should work (you probably want to modify it a bit to detect the highest number).
But I wouldn’t call it a “sort” anymore, the common name for this operation is “rotate”.
My advice would be to split it up into two function accordingly:

local count = find_highest_index(a)
rotate_left(a, count)
#10

no number - number is player id

And I need to sort the characters so that they are not on the same line.

active character always on top. Whoever goes next is behind him and so on

#12

The following function maps these:

local function cycle_table(t)
  for i=1,#t do
    if t[i][2]==1 then
      local n_t={}
      for ii=1,#t do
        n_t[ii]=t[i+ii-2<#t and i+ii-1 or i+ii-1-#t]
      end
      return n_t
    end
  end
end
-- usage:
xx = cycle_table(x)
#13

Thank you. Works. I can’t get used to random tables