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).
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);
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);
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
NewValue = (((OldValue - OldMin) * (NewMax - NewMin)) / (OldMax - OldMin)) + NewMin
或者更容易读懂:
OldRange = (OldMax - OldMin)
NewRange = (NewMax - NewMin)
NewValue = (((OldValue - OldMin) * NewRange) / OldRange) + NewMin
或者如果你想保护旧范围为0的情况(OldMin = OldMax):
OldRange = (OldMax - OldMin)
if (OldRange == 0)
NewValue = NewMin
else
{
NewRange = (NewMax - NewMin)
NewValue = (((OldValue - OldMin) * NewRange) / OldRange) + NewMin
}
注意，在这种情况下，我们被迫任意选择一个可能的新范围值。根据上下文，明智的选择可能是:NewMin(见示例)，NewMax或(NewMin + NewMax) / 2
使用Numpy和interp函数，你可以将你的值从旧范围转换为新范围:
>>> import numpy as np
>>> np.interp(0, [-16000,16000], [0,100])
50.0
你也可以尝试映射一个值列表:
>>> np.interp([-16000,0,12000] ,[-16000,16000], [0,100])
array([ 0. , 50. , 87.5])
