Convert number from old range to new numeric range

Question

I have a function that takes a specific number and a range, and re-scales it to the corresponding number in a different range.

/**
 * @param   int     number      The number to adjust.
 * @param   int     oldMin      The old minimum value.
 * @param   int     oldMax      The old maximum value.
 * @param   int     newMin      The new minimum value to convert to.
 * @param   int     newMax      The new maximium value to convert to.
 * 
 * @returns int     The original number converted to the new boundary.
*/
function convertValueToBoundary(number, oldMin, oldMax, newMin, newMax)
{
    return (((number - oldMin) * (newMax - newMin)) / (oldMax - oldMin)) + newMin
}

Formula

newValue = (((oldValue - oldMinimum) * (newMaximum - newMinimum)) / (oldMaximum - oldMinimum)) + newMinimum

Example

For instance, given the number 5 within a range of 0-10, we want to increase the numerical range to be between 0-100 instead.

convertValueToBoundary(5, 0, 10, 0, 100) // Output -> 50

Optimization & Improvements

Is this the optimal way to conduct this calculation? After a number of different attempts the above formula is the best I could come up with. Given a scenario where this needs to execute multiple times in rapid succession, it would be ideal to have this execute with the least amount of arithmetic operations required.

The conversion should be conducted without the use of any native language functions as I plan to use the same formula across different languages and platforms.

Answer 1

Code noise

Your naming and code style make the code very noisy.

The function name is too long convertValueToBoundary the conversion is implied and value is commonly shortened to val so valToBoundary is less noisy. However I would never guess what this function did without some extra information

Maybe rescale would be more appropriate.

You have too many brackets. You have...

return (((number - oldMin) * (newMax - newMin)) / (oldMax - oldMin)) + newMin

is identical to...

return (number - oldMin) * (newMax - newMin) / (oldMax - oldMin) + newMin;

To further remove code noise use an arrow function and shorten the argument names

const rescale = (val, oMin, oMax, min, max)=>(val - oMin) * (max - min) / (oMax - oMin) + min;

Optimization

The only optimization possible is to remove the division (multiplication is quicker than division). This would improve the performance if you were converting many values from the same range...

const rescale = (val, oMin, invRange, min, max)=>(val - oMin) * (max - min) * invRange + min;

// Your example convertValueToBoundary(5, 0, 10, 0, 100) // Output -> 50
// would be
rescale(5, 0, 0.1, 0, 100); // result 50;  is quicker
// or
rescale(5, 0, 1 / 10, 0, 100); // result 50; is slower as need the / and *

However if each range is different this would make the operation slower as you need to calculate the inverse range each time you call the function.

You could also use two ranges to remove the subtraction making it even quicker

const rescale = (val, oMin, invRange, min, range)=>(val - oMin) * range * invRange + min;

rescale(5, 0, 0.1, 0, 100); // result 50

Guard

If you keep the function with the division you need to guard against divide by zero. In JS convertValueToBoundary(5, 0, 0, 0, 100) will result in NaN rather than the expected 0 and in some languages it will throw an error.

Thus you should have the guard oMin === oMax to prevent the divide by zero.

const rescale = (val, oMin, oMax, min, max) => 
    (oMin === oMax ? 0 : (val - oMin) * (max - min) / (oMax - oMin)) + min;

If you use inverse range there is no need for the guard.

const rescale = (val, oMin, invRange, min, max)=>(val - oMin) * (max - min) * invRange + min;

// or

const rescale = (val, oMin, invRange, min, range)=>(val - oMin) * range * invRange + min;