Convert a number range to another range, maintaini

Question

I'm trying to convert one range of numbers to another, maintaining ratio. Maths is not my strong point.

I have an image file where point values may range from -16000.00 to 16000.00 though the typical range may be much less. What I want to do is compress these values into the integer range 0-100, where 0 is the value of the smallest point, and 100 is the value of the largest. All points in between should keep a relative ratio even though some precision is being lost I'd like to do this in python but even a general algorithm should suffice. I'd prefer an algorithm where the min/max or either range can be adjusted (ie, the second range could be -50 to 800 instead of 0 to 100).

Answer 1

That's a simple linear conversion.

new_value = ( (old_value - old_min) / (old_max - old_min) ) * (new_max - new_min) + new_min

So converting 10000 on the scale of -16000 to 16000 to a new scale of 0 to 100 yields:

old_value = 10000
old_min = -16000
old_max = 16000
new_min = 0
new_max = 100

new_value = ( ( 10000 - -16000 ) / (16000 - -16000) ) * (100 - 0) + 0
          = 81.25

Answer 2

This example converts a songs current position into an angle range of 20 - 40.

    /// <summary>
    /// This test converts Current songtime to an angle in a range. 
    /// </summary>
    [Fact]
    public void ConvertRangeTests()
    {            
       //Convert a songs time to an angle of a range 20 - 40
        var result = ConvertAndGetCurrentValueOfRange(
            TimeSpan.Zero, TimeSpan.FromMinutes(5.4),
            20, 40, 
            2.7
            );

        Assert.True(result == 30);
    }

    /// <summary>
    /// Gets the current value from the mixValue maxValue range.        
    /// </summary>
    /// <param name="startTime">Start of the song</param>
    /// <param name="duration"></param>
    /// <param name="minValue"></param>
    /// <param name="maxValue"></param>
    /// <param name="value">Current time</param>
    /// <returns></returns>
    public double ConvertAndGetCurrentValueOfRange(
                TimeSpan startTime,
                TimeSpan duration,
                double minValue,
                double maxValue,
                double value)
    {
        var timeRange = duration - startTime;
        var newRange = maxValue - minValue;
        var ratio = newRange / timeRange.TotalMinutes;
        var newValue = value * ratio;
        var currentValue= newValue + minValue;
        return currentValue;
    }

Answer 3

In the listing provided by PenguinTD, I do not understand why the ranges are reversed, it works without having to reverse the ranges. Linear range conversion is based upon the linear equation Y=Xm+n, where m and n are derived from the given ranges. Rather than refer to the ranges as min and max, it would be better to refer to them as 1 and 2. So the formula would be:

Y = (((X - x1) * (y2 - y1)) / (x2 - x1)) + y1

Where Y=y1 when X=x1, and Y=y2 when X=x2. x1, x2, y1 & y2 can be given any positive or negative value. Defining the expression in a macro makes it more useful,it can then be used with any argument names.

#define RangeConv(X, x1, x2, y1, y2) (((float)((X - x1) * (y2 - y1)) / (x2 - x1)) + y1)

The float cast would ensure floating point division in the case where all the arguments are integer values. Depending on the application it may not be necessary to check the ranges x1=x2 and y1==y2.

Answer 4

There is a condition, when all of the values that you are checking are the same, where @jerryjvl's code would return NaN.

if (OldMin != OldMax && NewMin != NewMax):
    return (((OldValue - OldMin) * (NewMax - NewMin)) / (OldMax - OldMin)) + NewMin
else:
    return (NewMax + NewMin) / 2

Answer 5

NewValue = (((OldValue - OldMin) * (NewMax - NewMin)) / (OldMax - OldMin)) + NewMin

Or a little more readable:

OldRange = (OldMax - OldMin)  
NewRange = (NewMax - NewMin)  
NewValue = (((OldValue - OldMin) * NewRange) / OldRange) + NewMin

Or if you want to protect for the case where the old range is 0 (OldMin = OldMax):

OldRange = (OldMax - OldMin)
if (OldRange == 0)
    NewValue = NewMin
else
{
    NewRange = (NewMax - NewMin)  
    NewValue = (((OldValue - OldMin) * NewRange) / OldRange) + NewMin
}

Note that in this case we're forced to pick one of the possible new range values arbitrarily. Depending on context, sensible choices could be: NewMin (see sample), NewMax or (NewMin + NewMax) / 2

Answer 6

I used this solution in a problem I was solving in js, so I thought I would share the translation. Thanks for the explanation and solution.

function remap( x, oMin, oMax, nMin, nMax ){
//range check
if (oMin == oMax){
    console.log("Warning: Zero input range");
    return None;
};

if (nMin == nMax){
    console.log("Warning: Zero output range");
    return None
}

//check reversed input range
var reverseInput = false;
oldMin = Math.min( oMin, oMax );
oldMax = Math.max( oMin, oMax );
if (oldMin != oMin){
    reverseInput = true;
}

//check reversed output range
var reverseOutput = false;  
newMin = Math.min( nMin, nMax )
newMax = Math.max( nMin, nMax )
if (newMin != nMin){
    reverseOutput = true;
};

var portion = (x-oldMin)*(newMax-newMin)/(oldMax-oldMin)
if (reverseInput){
    portion = (oldMax-x)*(newMax-newMin)/(oldMax-oldMin);
};

var result = portion + newMin
if (reverseOutput){
    result = newMax - portion;
}

return result;
}