我试图将一个范围的数字转换为另一个,保持比率。数学不是我的强项。

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).


当前回答

我没有为此挖掘BNF,但Arduino文档有一个很好的函数示例,它是分解的。我可以在Python中通过简单地添加一个def重命名到remap(因为map是内置的)并删除类型强制转换和花括号(即删除所有的'long')来使用它。

原始

long map(long x, long in_min, long in_max, long out_min, long out_max)
{
  return (x - in_min) * (out_max - out_min) / (in_max - in_min) + out_min;
}

Python

def remap(x, in_min, in_max, out_min, out_max):
  return (x - in_min) * (out_max - out_min) / (in_max - in_min) + out_min

https://www.arduino.cc/en/reference/map

其他回答

在由PenguinTD提供的清单中,我不明白为什么范围是颠倒的,它不需要颠倒范围就能工作。线性范围转换基于线性方程Y=Xm+n,其中m和n是从给定的范围推导出来的。与其将范围称为min和max,不如将它们称为1和2。所以公式是:

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

当X=x1时Y=y1,当X=x2时Y=y2。X1, x2, y1和y2可以取任意正值或负值。在宏中定义表达式使其更有用,它可以与任何参数名称一起使用。

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

在所有实参都是整数值的情况下,浮点强制转换将确保浮点除法。 根据应用程序的不同,可能不需要检查x1=x2和y1==y2的范围。

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;
    }

我个人使用支持泛型的helper类(Swift 3,4)。x兼容)

struct Rescale<Type : BinaryFloatingPoint> {
    typealias RescaleDomain = (lowerBound: Type, upperBound: Type)

    var fromDomain: RescaleDomain
    var toDomain: RescaleDomain

    init(from: RescaleDomain, to: RescaleDomain) {
        self.fromDomain = from
        self.toDomain = to
    }

    func interpolate(_ x: Type ) -> Type {
        return self.toDomain.lowerBound * (1 - x) + self.toDomain.upperBound * x;
    }

    func uninterpolate(_ x: Type) -> Type {
        let b = (self.fromDomain.upperBound - self.fromDomain.lowerBound) != 0 ? self.fromDomain.upperBound - self.fromDomain.lowerBound : 1 / self.fromDomain.upperBound;
        return (x - self.fromDomain.lowerBound) / b
    }

    func rescale(_ x: Type )  -> Type {
        return interpolate( uninterpolate(x) )
    }
}

Ex:

   let rescaler = Rescale<Float>(from: (-1, 1), to: (0, 100))
    
   print(rescaler.rescale(0)) // OUTPUT: 50

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;
}

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; }