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).
PHP的港口
发现PenguinTD的解决方案很有用，所以我将其移植到PHP。帮助你自己!
/**
* =====================================
* Remap Range
* =====================================
* - Convert one range to another. (including value)
*
* @param int $intValue The value in the old range you wish to convert
* @param int $oMin The minimum of the old range
* @param int $oMax The maximum of the old range
* @param int $nMin The minimum of the new range
* @param int $nMax The maximum of the new range
*
* @return float $fResult The old value converted to the new range
*/
function remapRange($intValue, $oMin, $oMax, $nMin, $nMax) {
// Range check
if ($oMin == $oMax) {
echo 'Warning: Zero input range';
return false;
}
if ($nMin == $nMax) {
echo 'Warning: Zero output range';
return false;
}
// Check reversed input range
$bReverseInput = false;
$intOldMin = min($oMin, $oMax);
$intOldMax = max($oMin, $oMax);
if ($intOldMin != $oMin) {
$bReverseInput = true;
}
// Check reversed output range
$bReverseOutput = false;
$intNewMin = min($nMin, $nMax);
$intNewMax = max($nMin, $nMax);
if ($intNewMin != $nMin) {
$bReverseOutput = true;
}
$fRatio = ($intValue - $intOldMin) * ($intNewMax - $intNewMin) / ($intOldMax - $intOldMin);
if ($bReverseInput) {
$fRatio = ($intOldMax - $intValue) * ($intNewMax - $intNewMin) / ($intOldMax - $intOldMin);
}
$fResult = $fRatio + $intNewMin;
if ($bReverseOutput) {
$fResult = $intNewMax - $fRatio;
}
return $fResult;
}
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
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)
}
}
PHP的港口
发现PenguinTD的解决方案很有用，所以我将其移植到PHP。帮助你自己!
/**
* =====================================
* Remap Range
* =====================================
* - Convert one range to another. (including value)
*
* @param int $intValue The value in the old range you wish to convert
* @param int $oMin The minimum of the old range
* @param int $oMax The maximum of the old range
* @param int $nMin The minimum of the new range
* @param int $nMax The maximum of the new range
*
* @return float $fResult The old value converted to the new range
*/
function remapRange($intValue, $oMin, $oMax, $nMin, $nMax) {
// Range check
if ($oMin == $oMax) {
echo 'Warning: Zero input range';
return false;
}
if ($nMin == $nMax) {
echo 'Warning: Zero output range';
return false;
}
// Check reversed input range
$bReverseInput = false;
$intOldMin = min($oMin, $oMax);
$intOldMax = max($oMin, $oMax);
if ($intOldMin != $oMin) {
$bReverseInput = true;
}
// Check reversed output range
$bReverseOutput = false;
$intNewMin = min($nMin, $nMax);
$intNewMax = max($nMin, $nMax);
if ($intNewMin != $nMin) {
$bReverseOutput = true;
}
$fRatio = ($intValue - $intOldMin) * ($intNewMax - $intNewMin) / ($intOldMax - $intOldMin);
if ($bReverseInput) {
$fRatio = ($intOldMax - $intValue) * ($intNewMax - $intNewMin) / ($intOldMax - $intOldMin);
}
$fResult = $fRatio + $intNewMin;
if ($bReverseOutput) {
$fResult = $intNewMax - $fRatio;
}
return $fResult;
}
