增加了KOTLIN版本的数学解释

假设我们有一个介于(OMin, Omax)之间的刻度，我们在这个范围内有一个值X

我们要把它转换成比例(NMin, NMax)

我们知道X，我们需要找到Y，比值必须相等:

 => (Y-NMin)/(NMax-NMin) = (X-OMin)/(OMax-OMin)  
      
 =>  (Y-NMin)/NewRange = (X-OMin)/OldRange 

 =>   Y = ((X-OMin)*NewRange)/oldRange)+NMin  Answer
   

从实用主义的角度来看，我们可以这样写这个问句:

 private fun  convertScale(oldValueToConvert:Int): Float {
       // Old Scale 50-100
       val oldScaleMin = 50
       val oldScaleMax = 100
       val oldScaleRange= (oldScaleMax - oldScaleMin)

       //new Scale 0-1
       val newScaleMin = 0.0f
       val newScaleMax = 1.0f
       val newScaleRange=  (newScaleMax - newScaleMin)
     
       return ((oldValueToConvert - oldScaleMin)* newScaleRange/ oldScaleRange) + newScaleMin
    }

JAVA

/**
     * 
     * @param x
     * @param inMin
     * @param inMax
     * @param outMin
     * @param outMax
     * @return
     */
        private long normalize(long x, long inMin, long inMax, long outMin, long outMax) {
          long outRange = outMax - outMin;
          long inRange  = inMax - inMin;
          return (x - inMin) *outRange / inRange + outMin;
        }

用法:

float brightness = normalize(progress, 0, 10, 0,255);

这个例子将歌曲的当前位置转换为20 - 40的角度范围。

    /// <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;
    }

C++变体

我发现PenguinTD的解决方案很有用，所以我把它移植到c++，如果有人需要它:

float remap(float x, float oMin, float oMax, float nMin, float nMax ){ //range check if( oMin == oMax) { //std::cout<< "Warning: Zero input range"; return -1; } if( nMin == nMax){ //std::cout<<"Warning: Zero output range"; return -1; } //check reversed input range bool reverseInput = false; float oldMin = min( oMin, oMax ); float oldMax = max( oMin, oMax ); if (oldMin == oMin) reverseInput = true; //check reversed output range bool reverseOutput = false; float newMin = min( nMin, nMax ); float newMax = max( nMin, nMax ); if (newMin == nMin) reverseOutput = true; float portion = (x-oldMin)*(newMax-newMin)/(oldMax-oldMin); if (reverseInput) portion = (oldMax-x)*(newMax-newMin)/(oldMax-oldMin); float result = portion + newMin; if (reverseOutput) result = newMax - portion; return result; }

我在一个用js解决的问题中使用了这个解决方案，所以我想我将分享翻译。谢谢你的解释和解决方案。

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;
}

我写了一个函数用R来做这个，方法和上面一样，但是我需要在R中做很多次，所以我想分享一下，以防它对任何人有帮助。

convertRange <- function(
  oldValue,
  oldRange = c(-16000.00, 16000.00), 
  newRange = c(0, 100),
  returnInt = TRUE # the poster asked for an integer, so this is an option
){
  oldMin <- oldRange[1]
  oldMax <- oldRange[2]
  newMin <- newRange[1]
  newMax <- newRange[2]
  newValue = (((oldValue - oldMin)* (newMax - newMin)) / (oldMax - oldMin)) + newMin
  
  if(returnInt){
   return(round(newValue))
  } else {
   return(newValue)
  }
}

实际上，在某些情况下，上述答案会失效。 如错误的输入值，错误的输入范围，负输入/输出范围。

def remap( x, oMin, oMax, nMin, nMax ):

    #range check
    if oMin == oMax:
        print "Warning: Zero input range"
        return None

    if nMin == nMax:
        print "Warning: Zero output range"
        return None

    #check reversed input range
    reverseInput = False
    oldMin = min( oMin, oMax )
    oldMax = max( oMin, oMax )
    if not oldMin == oMin:
        reverseInput = True

    #check reversed output range
    reverseOutput = False   
    newMin = min( nMin, nMax )
    newMax = max( nMin, nMax )
    if not newMin == nMin :
        reverseOutput = True

    portion = (x-oldMin)*(newMax-newMin)/(oldMax-oldMin)
    if reverseInput:
        portion = (oldMax-x)*(newMax-newMin)/(oldMax-oldMin)

    result = portion + newMin
    if reverseOutput:
        result = newMax - portion

    return result

#test cases
print remap( 25.0, 0.0, 100.0, 1.0, -1.0 ), "==", 0.5
print remap( 25.0, 100.0, -100.0, -1.0, 1.0 ), "==", -0.25
print remap( -125.0, -100.0, -200.0, 1.0, -1.0 ), "==", 0.5
print remap( -125.0, -200.0, -100.0, -1.0, 1.0 ), "==", 0.5
#even when value is out of bound
print remap( -20.0, 0.0, 100.0, 0.0, 1.0 ), "==", -0.2