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

列出理解一行的解决方案

color_array_new = [int((((x - min(node_sizes)) * 99) / (max(node_sizes) - min(node_sizes))) + 1) for x in node_sizes]

完整版

def colour_specter(waste_amount):
color_array = []
OldRange = max(waste_amount) - min(waste_amount)
NewRange = 99
for number_value in waste_amount:
    NewValue = int((((number_value - min(waste_amount)) * NewRange) / OldRange) + 1)
    color_array.append(NewValue)
print(color_array)
return color_array

其他回答

列出理解一行的解决方案

color_array_new = [int((((x - min(node_sizes)) * 99) / (max(node_sizes) - min(node_sizes))) + 1) for x in node_sizes]

完整版

def colour_specter(waste_amount):
color_array = []
OldRange = max(waste_amount) - min(waste_amount)
NewRange = 99
for number_value in waste_amount:
    NewValue = int((((number_value - min(waste_amount)) * NewRange) / OldRange) + 1)
    color_array.append(NewValue)
print(color_array)
return color_array

Java版本

不管你喂它什么,它都能工作!

我把所有内容都展开了,这样便于学习。当然,最后舍入是可选的。

    private long remap(long p, long Amin, long Amax, long Bmin, long Bmax ) {

    double deltaA = Amax - Amin;
    double deltaB = Bmax - Bmin;
    double scale  = deltaB / deltaA;
    double negA   = -1 * Amin;
    double offset = (negA * scale) + Bmin;
    double q      = (p * scale) + offset;
    return Math.round(q);

}

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;
    }

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。