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;
}
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);
下面是一些简单的Python函数，便于复制和粘贴，包括一个扩展整个列表的函数。
def scale_number(unscaled, to_min, to_max, from_min, from_max):
return (to_max-to_min)*(unscaled-from_min)/(from_max-from_min)+to_min
def scale_list(l, to_min, to_max):
return [scale_number(i, to_min, to_max, min(l), max(l)) for i in l]
可以这样使用:
scale_list([1,3,4,5], 0, 100)
[0.0, 50.0, 75.0, 100.0]
在我的例子中，我想缩放一条对数曲线，像这样:
scale_list([math.log(i+1) for i in range(5)], 0, 50)
[0.0, 21.533827903669653, 34.130309724299266, 43.06765580733931, 50.0]
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])
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
