Simple Round

wiki

function DIV(a,b)
    return (a - a % b) / b
end

to get an integer result during a division. (JlnWntr)

The following function rounds a number to the given number of decimal places.

function round(num, numDecimalPlaces)
  local mult = 10^(numDecimalPlaces or 0)
  return math.floor(num * mult + 0.5) / mult
end

Here is an alternative implementation:

function round2(num, numDecimalPlaces)
  return tonumber(string.format("%." .. (numDecimalPlaces or 0) .. "f", num))
end

Both are Lua 5.0 and 5.1 compatible.

If the number is rounded in order to be printed, then remove the tonumber: Converting to number then back to string would reintroduce rounding errors.

Tests:

> function test(a, b) print(round(a,b), round2(a,b)) end
> test(43245325.9995, 3)
43245326        43245325.999
> test(43245325.9994, 3)
43245325.999    43245325.999
> test(43245325.5543654)
43245326        43245326
> test(43245325.5543654, 3)
43245325.554    43245325.554
> test(43245325.5543654, 4)
43245325.5544   43245325.5544

Note: The first function will misbehave if numDecimalPlaces is negative, so this version might be more robust (Robert Jay Gould)

round(1023.4345, -2) = 1000
round(1023.4345, 2) = 1023.43

function round(num, numDecimalPlaces)
  if numDecimalPlaces and numDecimalPlaces>0 then
    local mult = 10^numDecimalPlaces
    return math.floor(num * mult + 0.5) / mult
  end
  return math.floor(num + 0.5)
end

function round(num) return math.floor(num+.5) end

Might have unintended result for -.5? Please see below.

-- Henning

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

Note that math.ceil(num-.5) ~= math.floor(num+.5) e.g. for -.5 with math.ceil(num-.5) == -1 math.floor(num+.5) == 0

Samples: 1.1 -> 1, 1 -> 1, 0.9 -> 1, 0.5 -> 1, 0.1 -> 0, 0 -> 0, -0.1 -> 0, -0.4 -> 0, -0.5 -> -1, -0.6 -> -1, -0.9 -> -1, -1: -1, -1.1: -1

-- Henning

function round(num)
    under = math.floor(num)
    upper = math.floor(num) + 1
    underV = -(under - num)
    upperV = upper - num
    if (upperV > underV) then
        return under
    else
        return upper
    end
end

function round(num, numDecimalPlaces)
    local mult = 10^(numDecimalPlaces or 0)
    if num >= 0 then return math.floor(num * mult + 0.5) / mult
    else return math.ceil(num * mult - 0.5) / mult end
end

-- Igor Skoric (i.skoric@student.tugraz.at)

function round(n, mult)
    mult = mult or 1
    return math.floor((n + mult/2)/mult) * mult
end

--should round and give negatives too if I'm not mistaken
function round(num, numDecimalPlaces)
  if numDecimalPlaces and numDecimalPlaces>0 then
    local mult = 10^numDecimalPlaces
    return math.floor(num * mult + 0.5) / mult
  else
    numDecimalPlaces = numDecimalPlaces mult 2 -- negates its negative status
    local mult = 10^numDecimalPlaces
    return math.floor(num * mult + 0.5) / mult
    ide = numDecimalPlaces / (numDecimalPlaces/2)
  end
  return math.floor(num + 0.5)
end

Math.modf is your friend!

if num<0 then x=-.5 else x=.5 end
Integer, decimal = math.modf(num+x)

For rounding to a decimal place I use

roundto=10
if num*roundto<0 then x=-.5 else x=.5 end
Integer, decimal = math.modf(num*roundto+x)
result = Integer/roundto

result will equal number rounded to the decimal place "roundto"

function round(n)
  return math.floor((math.floor(n*2) + 1)/2)
end

function math.sign(v)
	return (v >= 0 and 1) or -1
end
function math.round(v, bracket)
	bracket = bracket or 1
	return math.floor(v/bracket + math.sign(v) * 0.5) * bracket
end

math.round(119.68, 6.4) -- 121.6 (= 19 * 6.4)

math.round(119.68, 0.01) -- 119.68
math.round(119.68, 0.1) -- 119.7
math.round(119.68) -- 120
math.round(119.68, 100) -- 100
math.round(119.68, 1000) -- 0

Who's going to make the next round function, I wonder? -- Lucids