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).
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]
我写了一个函数用R来做这个，方法和上面一样，但是我需要在R中做很多次，所以我想分享一下，以防它对任何人有帮助。
convertRange <- function(
oldValue,
oldRange = c(-16000.00, 16000.00),
newRange = c(0, 100),
returnInt = TRUE # the poster asked for an integer, so this is an option
){
oldMin <- oldRange[1]
oldMax <- oldRange[2]
newMin <- newRange[1]
newMax <- newRange[2]
newValue = (((oldValue - oldMin)* (newMax - newMin)) / (oldMax - oldMin)) + newMin
if(returnInt){
return(round(newValue))
} else {
return(newValue)
}
}
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
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);
}
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);
我没有为此挖掘BNF，但Arduino文档有一个很好的函数示例，它是分解的。我可以在Python中通过简单地添加一个def重命名到remap(因为map是内置的)并删除类型强制转换和花括号(即删除所有的'long')来使用它。
原始
long map(long x, long in_min, long in_max, long out_min, long out_max)
{
return (x - in_min) * (out_max - out_min) / (in_max - in_min) + out_min;
}
Python
def remap(x, in_min, in_max, out_min, out_max):
return (x - in_min) * (out_max - out_min) / (in_max - in_min) + out_min
https://www.arduino.cc/en/reference/map
