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).
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])
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。
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);
}
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