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).


当前回答

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;
}

其他回答

实际上,在某些情况下,上述答案会失效。 如错误的输入值,错误的输入范围,负输入/输出范围。

def remap( x, oMin, oMax, nMin, nMax ):

    #range check
    if oMin == oMax:
        print "Warning: Zero input range"
        return None

    if nMin == nMax:
        print "Warning: Zero output range"
        return None

    #check reversed input range
    reverseInput = False
    oldMin = min( oMin, oMax )
    oldMax = max( oMin, oMax )
    if not oldMin == oMin:
        reverseInput = True

    #check reversed output range
    reverseOutput = False   
    newMin = min( nMin, nMax )
    newMax = max( nMin, nMax )
    if not newMin == nMin :
        reverseOutput = True

    portion = (x-oldMin)*(newMax-newMin)/(oldMax-oldMin)
    if reverseInput:
        portion = (oldMax-x)*(newMax-newMin)/(oldMax-oldMin)

    result = portion + newMin
    if reverseOutput:
        result = newMax - portion

    return result

#test cases
print remap( 25.0, 0.0, 100.0, 1.0, -1.0 ), "==", 0.5
print remap( 25.0, 100.0, -100.0, -1.0, 1.0 ), "==", -0.25
print remap( -125.0, -100.0, -200.0, 1.0, -1.0 ), "==", 0.5
print remap( -125.0, -200.0, -100.0, -1.0, 1.0 ), "==", 0.5
#even when value is out of bound
print remap( -20.0, 0.0, 100.0, 0.0, 1.0 ), "==", -0.2

这是一个简单的线性变换。

new_value = ( (old_value - old_min) / (old_max - old_min) ) * (new_max - new_min) + new_min

因此,将10000在-16000到16000的范围内转换为0到100的新范围会得到:

old_value = 10000
old_min = -16000
old_max = 16000
new_min = 0
new_max = 100

new_value = ( ( 10000 - -16000 ) / (16000 - -16000) ) * (100 - 0) + 0
          = 81.25

列出理解一行的解决方案

color_array_new = [int((((x - min(node_sizes)) * 99) / (max(node_sizes) - min(node_sizes))) + 1) for x in node_sizes]

完整版

def colour_specter(waste_amount):
color_array = []
OldRange = max(waste_amount) - min(waste_amount)
NewRange = 99
for number_value in waste_amount:
    NewValue = int((((number_value - min(waste_amount)) * NewRange) / OldRange) + 1)
    color_array.append(NewValue)
print(color_array)
return color_array

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);