lua table find higest value

[benanderson_475]

I want to get the highest time and then switch the groups on in the order from highest time(stored in on_time) to lowest time until all the groups are on, groups/ range i am interested in are 1/1/1 to 1/1/20
i am storing the obj.updatetime in a table with the associated address etc as below

Code:

function chk_off()
  t ={}

for i = 1, 20 do
    addr = '1/1/'.. tostring(i)
    obj = grp.find(addr)
   
  if obj and obj.value == 0  then
     --put all off obj in table with update time
      table.insert(t, {address = addr, level = obj.value, upd_time = obj.updatetime, on_time = os.time() -obj.updatetime }  )
    end     
end
  return t
end 

Then i want to get the highest value from all the t.on_time/s and switch each off, in order from the longest off value (stored in on_time) until all the complete range 1/1/1 - 1/1/20 are all off
this runs from a function which is called from a scheduled script every x minutes.
This is where i am stuck, this is what i have so far, my question is how can i calculate the highest on_time value and what will happen if two possibility have the same on_time value?
I hope this is making sense....

Code:

function load_time_off()
t = chk_off()
if t then  
for k, v in ipairs(t) do
log(v.on_time) -- this gives me all the on_time vals
-- how to get highest on_time value?? and switch on the associated address
   end
end    
end

[admin]

You can find the maximum value like this. If two items have the same on_time then the one that appears first in the table will be returned.

Code:

max_index = 0
max_time = 0

for index, item in ipairs(t) do
  if item.on_time > max_time then
    max_index = index
    max_time = item.on_time
  end
end

max_item = t[ max_index ]
if max_item then
  log(max_item)
end

Or you can sort the table using custom comparison function. In this case the order of elements in not defined if on_time is the same. t[1] will contain the element with maximum on_time value.

Code:

table.sort(t, function(a, b)
  return a.on_time > b.on_time
end)

log(t)

[benanderson_475]

You can find the maximum value like this. If two items have the same on_time then the one that appears first in the table will be returned.

Code:

max_index = 0
max_time = 0

for index, item in ipairs(t) do
  if item.on_time > max_time then
    max_index = index
    max_time = item.on_time
  end
end

max_item = t[ max_index ]
if max_item then
  log(max_item)
end

Or you can sort the table using custom comparison function. In this case the order of elements in not defined if on_time is the same. t[1] will contain the element with maximum on_time value.

Code:

table.sort(t, function(a, b)
  return a.on_time > b.on_time
end)

log(t)

Many Thanks, i have it working

[manos@dynamitec]

Hello Admin,

I am using the following to find max value of multiple tables with multiple tagged objects. Now I am repeating the same calculation many times in a resident script; how can I achieve the same but with something like a table containing multiple tables in and run the for loop once per table?

Code:

--Actual Lux Values
lux_rest = grp.tag('Lux_Rest')
lux_hal1 = grp.tag('Lux_H1')
lux_hal2 = grp.tag('Lux_H2')
lux_hal3 = grp.tag('Lux_H3')
lux_hal4 = grp.tag('Lux_H4')
lux_hal5 = grp.tag('Lux_H5')
lux_hal6 = grp.tag('Lux_H6')
lux_hal7 = grp.tag('Lux_H7')


-----------MAX LUX VALUE--------------
-- Calculate Max Lux Value in Zone Under Restaurant
t0 = {}
for i, v in ipairs(lux_rest) do
  table.insert(t0, v.value)
end

if t0 then
  table.sort(t0)
    local max = t0[#t0]
    grp.checkwrite('3/0/0', max, 1)
end

os.sleep(0.1)

-- Calculate Max Lux Value in Hal 1
t1 = {}
for i, v in ipairs(lux_hal1) do
  table.insert(t1, v.value)
end

if t1 then
  table.sort(t1)
    local max = t1[#t1]
    grp.checkwrite('3/0/10', max, 1)
end

os.sleep(0.1)

-- Calculate Max Lux Value in Hal 2
t2 = {}
for i, v in ipairs(lux_hal2) do
  table.insert(t2, v.value)
end

if t2 then
  table.sort(t2)
    local max = t2[#t2]
    grp.checkwrite('3/0/20', max, 1)
end

os.sleep(0.1)

-- Calculate Max Lux Value in Hal 3
t3 = {}
for i, v in ipairs(lux_hal3) do
  table.insert(t3, v.value)
end

if t3 then
  table.sort(t3)
    local max = t3[#t3]
    grp.checkwrite('3/0/30', max, 1)
end

os.sleep(0.1)

-- Calculate Max Lux Value in Hal 4
t4 = {}
for i, v in ipairs(lux_hal4) do
  table.insert(t4, v.value)
end

if t4 then
  table.sort(t4)
    local max = t4[#t4]
    grp.checkwrite('3/0/40', max, 1)
end

os.sleep(0.1)

-- Calculate Max Lux Value in Hal 5
t5 = {}
for i, v in ipairs(lux_hal5) do
  table.insert(t5, v.value)
end

if t5 then
  table.sort(t5)
    local max = t5[#t5]
    grp.checkwrite('3/0/50', max, 1)
end

os.sleep(0.1)

-- Calculate Max Lux Value in Hal 6
t6 = {}
for i, v in ipairs(lux_hal6) do
  table.insert(t6, v.value)
end

if t6 then
  table.sort(t6)
    local max = t6[#t6]
    grp.checkwrite('3/0/60', max, 1)
end

os.sleep(0.1)

-- Calculate Max Lux Value in Hal 7
t7 = {}
for i, v in ipairs(lux_hal7) do
  table.insert(t7, v.value)
end

if t7 then
  table.sort(t7)
    local max = t7[#t7]
    grp.checkwrite('3/0/70', max, 1)
end

Kind regards

[admin]

Try this:

Code:

tags = {
  { tag = 'tag1', output = '1/1/1' },
  { tag = 'tag2', output = '2/1/1' },
}

function tag_max(tag)
  local objects = grp.tag(tag)
  local result = -math.huge

  for _, object in ipairs(objects) do
    result = math.max(result, object.value)
  end

  return result
end

for _, t in ipairs(tags) do
  max = tag_max(t.tag)
  grp.checkwrite(t.output, max)

  os.sleep(0.1)
end

[manos@dynamitec]

Try this:

Code:

tags = {
  { tag = 'tag1', output = '1/1/1' },
  { tag = 'tag2', output = '2/1/1' },
}

function tag_max(tag)
  local objects = grp.tag(tag)
  local result = -math.huge

  for _, object in ipairs(objects) do
    result = math.max(result, object.value)
  end

  return result
end

for _, t in ipairs(tags) do
  max = tag_max(t.tag)
  grp.checkwrite(t.output, max)

  os.sleep(0.1)
end

Thank you Admin. Works as expected .

If I want to also calculate the min value in the same script and write on diff object?