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).
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;
}
下面是一个Javascript版本，它返回一个函数，对预定的源和目标范围进行重新缩放，最大限度地减少每次必须执行的计算量。
// This function returns a function bound to the
// min/max source & target ranges given.
// oMin, oMax = source
// nMin, nMax = dest.
function makeRangeMapper(oMin, oMax, nMin, nMax ){
//range check
if (oMin == oMax){
console.log("Warning: Zero input range");
return undefined;
};
if (nMin == nMax){
console.log("Warning: Zero output range");
return undefined
}
//check reversed input range
var reverseInput = false;
let oldMin = Math.min( oMin, oMax );
let oldMax = Math.max( oMin, oMax );
if (oldMin != oMin){
reverseInput = true;
}
//check reversed output range
var reverseOutput = false;
let newMin = Math.min( nMin, nMax )
let newMax = Math.max( nMin, nMax )
if (newMin != nMin){
reverseOutput = true;
}
// Hot-rod the most common case.
if (!reverseInput && !reverseOutput) {
let dNew = newMax-newMin;
let dOld = oldMax-oldMin;
return (x)=>{
return ((x-oldMin)* dNew / dOld) + newMin;
}
}
return (x)=>{
let portion;
if (reverseInput){
portion = (oldMax-x)*(newMax-newMin)/(oldMax-oldMin);
} else {
portion = (x-oldMin)*(newMax-newMin)/(oldMax-oldMin)
}
let result;
if (reverseOutput){
result = newMax - portion;
} else {
result = portion + newMin;
}
return result;
}
}
下面是一个使用该函数将0-1缩放到-0x80000000, 0x7FFFFFFF的示例
let normTo32Fn = makeRangeMapper(0, 1, -0x80000000, 0x7FFFFFFF);
let fs = normTo32Fn(0.5);
let fs2 = normTo32Fn(0);
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;
}
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);
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。
