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    Rounding

    HarrisonHarrison Member
    edited May 2014 in General questions
    In programming, rounding down to the ones place is important.. think hours to minutes, dollars to change.. etc.
    I made a exel spreadsheet that did just that-
    in psudocode:
x=5.223
y=rounddown(x) (to the nearest ones place)
x=x-y
print(x..":"..y)
    (in a perfect world would yield 5:0.22)
    but when i attempt to round in lua there is no simple round function unlike exel..
    math.floor just takes my number to zero D:
    so does this:
     function round(num) 
        if num >= 0 then return math.floor(num+.5) 
        else return math.ceil(num-.5) end
    end
    also yields 0
    which i got from http://lua-users.org/wiki/SimpleRound
    First, why?
    Next, what should i be doing?
    “ The first 90% of the code accounts for the first 90% of the development time. The remaining 10% of the code accounts for the other 90% of the development time. ” - Tom Cargill
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    Comments

    • amin13aamin13a Member
      It works for me:
      in main.lua (and no other code):

      function round(num)
      if num >= 0 then return math.floor(num+.5)
      else return math.ceil(num-.5) end
      end
      print(round(2.85))

      Result:
      main.lua is uploading.
      Uploading finished.
      3
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    • HubertRonaldHubertRonald Member
      edited May 2014
      Hi @Harrison

      I use the next code:
      local floor = math.floor
function round(val, decimal)
  if (decimal) then
    return floor( (val * 10^decimal) + 0.5) / (10^decimal)
  else
    return floor(val+0.5)
  end
end
 
 
print(math.pi)
print(round(math.pi))
print(round(math.pi,4))
 
--[[
   I get:
   3.1415926535898
   3
   3.1416
]]
      In your case I would do the following
      local x = 5.223 -- input
local y = round(x,2) -- define numbers of decimals
x = round(x) -- rounddown integer
y = y - x -- get decimals
print(table.concat({x,":",y})) -- it's more fast!
 
--[[
   I get:
   5:0.22
]]
      I hope than it can hep you :)>-

      [-] Liasoft
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    • HarrisonHarrison Member
      edited May 2014
      thanks, that helps a lot. Now i just need to fix the results when something with a decimal over 0.5 is passed (ex. 5.55 results in 6 and -0.45)
      edit:
      fixed by doing this after your second block of code you posted-
       if y < 0 then --checks to see if the rounding code rounded up 
y = 1 + y --takes the negative y and finds its positive opposite
x=x-1 --removes 1 from x to make up for upward round
end
      man this concept is hard to wrap my mind around! :D
      “ The first 90% of the code accounts for the first 90% of the development time. The remaining 10% of the code accounts for the other 90% of the development time. ” - Tom Cargill
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    • HubertRonaldHubertRonald Member
      @Harrison I'm glad you've got it :)>-

      [-] Liasoft
      +1 -1Share on Facebook
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