将一个数字范围转换为另一个范围，保持比率

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;
}

2016-03-27 06:37:22

其他回答

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

2009-05-30 05:50:45

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;
}

2014-07-09 16:14:08

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])

2021-09-21 04:53:17

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

2013-03-21 00:32:53

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;
}

2016-03-27 06:37:22

将一个数字范围转换为另一个范围，保持比率

推荐文章

最新文章

标签