我一直在编写一个图像处理程序,该程序通过HTML5 Canvas 像素处理来应用效果。我已经实现了阈值处理,Vintaging和ColorGradient像素操作,但是令人难以置信的是我无法更改图像的对比度!
我已经尝试了多种解决方案,但是我总是在图片中获得太多的亮度,而获得的对比度效果却很少,而且我不打算使用任何Javascript库,因为我试图在本机实现这些效果。

基本的像素操作代码:

var data = imageData.data;
for (var i = 0; i < data.length; i += 4) {
 //Note: data[i], data[i+1], data[i+2] represent RGB respectively
data[i] = data[i];
data[i+1] = data[i+1];
data[i+2] = data[i+2];
}

像素操纵示例

值处于RGB模式,这意味着data [i]是红色。因此,如果data [i] = data [i] * 2;该像素的红色通道的亮度将增加到两倍。例:
var data = imageData.data;
for (var i = 0; i < data.length; i += 4) {
 //Note: data[i], data[i+1], data[i+2] represent RGB respectively
 //Increases brightness of RGB channel by 2
data[i] = data[i]*2;
data[i+1] = data[i+1]*2;
data[i+2] = data[i+2]*2;
}

* 注意:我不是要你们完成代码!那将是一个忙!我要的是一种算法(甚至是伪代码),该算法显示像素操纵中的对比度是如何实现的!
如果有人可以为HTML5 canvas中的图像对比度提供良好的算法,我将感到非常高兴。

最佳答案

更快的选项(基于Escher's approach)是:

function contrastImage(imgData, contrast){  //input range [-100..100]
    var d = imgData.data;
    contrast = (contrast/100) + 1;  //convert to decimal & shift range: [0..2]
    var intercept = 128 * (1 - contrast);
    for(var i=0;i<d.length;i+=4){   //r,g,b,a
        d[i] = d[i]*contrast + intercept;
        d[i+1] = d[i+1]*contrast + intercept;
        d[i+2] = d[i+2]*contrast + intercept;
    }
    return imgData;
}

推导类似于以下内容;这个版本在数学上是相同的,但是运行速度要快得多。

原始答案

这是简化版,其中包含方法already discussed(基于this article)的解释:
function contrastImage(imageData, contrast) {  // contrast as an integer percent
    var data = imageData.data;  // original array modified, but canvas not updated
    contrast *= 2.55; // or *= 255 / 100; scale integer percent to full range
    var factor = (255 + contrast) / (255.01 - contrast);  //add .1 to avoid /0 error

    for(var i=0;i<data.length;i+=4)  //pixel values in 4-byte blocks (r,g,b,a)
    {
        data[i] = factor * (data[i] - 128) + 128;     //r value
        data[i+1] = factor * (data[i+1] - 128) + 128; //g value
        data[i+2] = factor * (data[i+2] - 128) + 128; //b value

    }
    return imageData;  //optional (e.g. for filter function chaining)
}

笔记
  • 我选择使用contrast+/- 100范围代替原始的+/- 255。对于不了解基本概念的用户或程序员而言,百分比值似乎更直观。另外,我的用法始终与UI控件相关;从-100%到+ 100%的范围允许我直接标记和绑定(bind)控制值,而不用进行调整或解释。
  • 即使calculated values can far exceed the allowable range,此算法也不包括范围检查-这是因为ImageData对象基础的数组是Uint8ClampedArrayAs MSDN explains,使用Uint8ClampedArray为您处理范围检查:



  • 用法

    请注意,尽管基本公式是相当对称的(允许往返),但由于像素仅允许整数值,因此在高级别过滤时会丢失数据。例如,当您将图像去饱和度达到极高的级别(> 95%左右)时,所有像素基本上都是均匀的中等灰度(在128个可能的平均值的几位之内)。再次将对比度调高会导致色彩范围变平。

    同样,在应用多个对比度调整时,操作顺序也很重要-饱和值会迅速“爆裂”(超出最大值255),这意味着高度饱和然后降低饱和度会导致整体图像变暗。然而,去饱和然后再饱和不会造成太多数据丢失,因为高光和阴影值被静音,而不是被裁剪(请参阅下面的说明)。

    一般来说,当应用多个滤镜时,最好从原始数据开始每个操作,然后依次重新应用每个调整,而不是尝试撤消以前的更改-至少对于图像质量而言。对于每种情况,性能速度或其他要求可能会有所不同。

    html - HTML5 Canvas图像对比度-LMLPHP

    代码示例:

    function contrastImage(imageData, contrast) {  // contrast input as percent; range [-1..1]
        var data = imageData.data;  // Note: original dataset modified directly!
        contrast *= 255;
        var factor = (contrast + 255) / (255.01 - contrast);  //add .1 to avoid /0 error.
    
        for(var i=0;i<data.length;i+=4)
        {
            data[i] = factor * (data[i] - 128) + 128;
            data[i+1] = factor * (data[i+1] - 128) + 128;
            data[i+2] = factor * (data[i+2] - 128) + 128;
        }
        return imageData;  //optional (e.g. for filter function chaining)
    }
    
    $(document).ready(function(){
      var ctxOrigMinus100 = document.getElementById('canvOrigMinus100').getContext("2d");
      var ctxOrigMinus50 = document.getElementById('canvOrigMinus50').getContext("2d");
      var ctxOrig = document.getElementById('canvOrig').getContext("2d");
      var ctxOrigPlus50 = document.getElementById('canvOrigPlus50').getContext("2d");
      var ctxOrigPlus100 = document.getElementById('canvOrigPlus100').getContext("2d");
    
      var ctxRoundMinus90 = document.getElementById('canvRoundMinus90').getContext("2d");
      var ctxRoundMinus50 = document.getElementById('canvRoundMinus50').getContext("2d");
      var ctxRound0 = document.getElementById('canvRound0').getContext("2d");
      var ctxRoundPlus50 = document.getElementById('canvRoundPlus50').getContext("2d");
      var ctxRoundPlus90 = document.getElementById('canvRoundPlus90').getContext("2d");
    
    
      var img = new Image();
      img.onload = function() {
        //draw orig
        ctxOrig.drawImage(img, 0, 0, img.width, img.height, 0, 0, 100, 100); //100 = canvas width, height
    
        //reduce contrast
        var origBits = ctxOrig.getImageData(0, 0, 100, 100);
        contrastImage(origBits, -.98);
        ctxOrigMinus100.putImageData(origBits, 0, 0);
    
        var origBits = ctxOrig.getImageData(0, 0, 100, 100);
        contrastImage(origBits, -.5);
        ctxOrigMinus50.putImageData(origBits, 0, 0);
    
        // add contrast
        var origBits = ctxOrig.getImageData(0, 0, 100, 100);
        contrastImage(origBits, .5);
        ctxOrigPlus50.putImageData(origBits, 0, 0);
    
        var origBits = ctxOrig.getImageData(0, 0, 100, 100);
        contrastImage(origBits, .98);
        ctxOrigPlus100.putImageData(origBits, 0, 0);
    
    
        //round-trip, de-saturate first
        origBits = ctxOrig.getImageData(0, 0, 100, 100);
        contrastImage(origBits, -.98);
        contrastImage(origBits, .98);
        ctxRoundMinus90.putImageData(origBits, 0, 0);
    
        origBits = ctxOrig.getImageData(0, 0, 100, 100);
        contrastImage(origBits, -.5);
        contrastImage(origBits, .5);
        ctxRoundMinus50.putImageData(origBits, 0, 0);
    
        //do nothing 100 times
        origBits = ctxOrig.getImageData(0, 0, 100, 100);
        for(i=0;i<100;i++){
          contrastImage(origBits, 0);
        }
        ctxRound0.putImageData(origBits, 0, 0);
    
        //round-trip, saturate first
        origBits = ctxOrig.getImageData(0, 0, 100, 100);
        contrastImage(origBits, .5);
        contrastImage(origBits, -.5);
        ctxRoundPlus50.putImageData(origBits, 0, 0);
    
        origBits = ctxOrig.getImageData(0, 0, 100, 100);
        contrastImage(origBits, .98);
        contrastImage(origBits, -.98);
        ctxRoundPlus90.putImageData(origBits, 0, 0);
      };
    
      img.src = 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";
    
    });
    canvas {width: 100px; height: 100px}
    div {text-align:center; width:120px; float:left}
    <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
    
    <div>
      <canvas id="canvOrigMinus100" width="100" height="100"></canvas>
      -98%
    </div>
    
    <div>
      <canvas id="canvOrigMinus50" width="100" height="100"></canvas>
      -50%
    </div>
    
    <div>
      <canvas id="canvOrig" width="100" height="100"></canvas>
      Original
    </div>
    
    <div>
      <canvas id="canvOrigPlus50" width="100" height="100"></canvas>
      +50%
    </div>
    
    <div>
      <canvas id="canvOrigPlus100" width="100" height="100"></canvas>
      +98%
    </div>
    
      <hr/>
    
    <div style="clear:left">
      <canvas id="canvRoundMinus90" width="100" height="100"></canvas>
      Round-trip <br/> (-98%, +98%)
    </div>
    
    <div>
      <canvas id="canvRoundMinus50" width="100" height="100"></canvas>
      Round-trip <br/> (-50%, +50%)
    </div>
    
    <div>
      <canvas id="canvRound0" width="100" height="100"></canvas>
      Round-trip <br/> (0% 100x)
    </div>
    
    <div>
      <canvas id="canvRoundPlus50" width="100" height="100"></canvas>
      Round-trip <br/> (+50%, -50%)
    </div>
    
    <div>
      <canvas id="canvRoundPlus90" width="100" height="100"></canvas>
      Round-trip <br/> (+98%, -98%)
    </div>


    说明

    (免责声明-我不是图像专家或数学家。我正在尝试提供具有最少技术细节的常识性解释。下面一些手动挥舞着手,例如255 = 256以避免索引问题,127.5 = 128以便简化数字。)

    因为对于给定的像素,颜色通道的非零值的可能数目为255 ,所以像素的“无对比度”,平均值为128 (或127,或127.5,如果您想,但差异可以忽略不计)。为了说明的目的,“对比度”的量是从当前值到平均值(128)的距离。调整对比度意味着增大或减小当前值和平均值之间的差异。

    该算法然后解决的问题是:
  • 选择一个常数因子以通过
  • 调整对比度
  • 对于每个像素的每个颜色通道,按该常数因子
  • 缩放“对比度”(与平均值的距离)

    或者,如CSS spec所示,只需选择直线的斜率和截距即可:



    注意术语type='linear';我们正在RGB color space中进行线性对比度调整,这与二次缩放函数,luminence-based调整或histogram matching相反。

    如果从几何类中回想起,则直线的公式为y=mx+by是我们要得到的最终值,坡度m是对比度(或factor),x是初始像素值,b是y轴的截距(x = 0),垂直移动该线。还请记住,由于y截距不在原点(0,0)处,所以该公式也可以表示为 y=m(x-a)+b ,其中a是水平偏移线的x偏移量。

    html - HTML5 Canvas图像对比度-LMLPHP

    就我们的目的而言,该图表示输入值(x轴)和结果(y轴)。我们已经知道,y截距b(对于m=0,没有对比度)必须为128(我们可以根据规格检查0.5-0.5 * 256的整个范围= 128)。 x是我们的原始值,因此我们需要找出斜率m和x-offset a

    首先,斜率m是“上升越过”或(y2-y1)/(x2-x1)-因此我们需要2个点已知在所需的线上。找到这些要点需要结合以下几点:
  • 我们的函数采用线截距图
  • 的形状
  • y截距为b = 128-不管斜率(对比度)如何。
  • 预期的最大“y”值为255,最小值为0
  • 可能的'x'值范围是256
  • 中性值应始终保持中性:128 => 128而不考虑斜率
  • 对比0的调整应导致输入和输出之间没有变化;即1:1的斜率。

  • 综合考虑所有这些因素,我们可以得出结论,无论施加了什么对比度(斜率),我们得到的线都将以128,128为中心(并绕其旋转)。由于我们的y截距非零,因此x截距也非零;我们知道x范围为256宽,并且居中居中,因此它必须偏移可能范围的一半:256/2 = 128。

    html - HTML5 Canvas图像对比度-LMLPHP

    因此,现在对于y=m(x-a)+b,我们知道m以外的所有信息。回想一下几何类中的两个要点:
  • 线即使位置发生变化也具有相同的斜率;也就是说,不管ma的值如何,b都保持不变。
  • 可以使用
  • 线上的任意2个点找到直线的斜率

    为了简化斜率讨论,让我们将坐标原点移动到x轴截距(-128),暂时忽略ab。现在,我们的原始行将绕过(0,0),我们知道该行的第二点位于(255,255)处的x(输入)和y(输出)的整个范围内。

    我们将使新线在(0,0)处旋转,因此可以将其用作新线上将遵循最终对比度斜率m的点之一。第二点可以通过将(255,255)处的当前端移动一定量来确定;由于我们仅限于单个输入(contrast)并使用线性函数,因此第二个点将在图形上的xy方向上平均移动。

    html - HTML5 Canvas图像对比度-LMLPHP

    4个可能的新点的(x,y)坐标将为255 +/- contrast。由于增加和减少x和y会使我们保持在原始的1:1行上,因此让我们看一下所示的+x, -y-x, +y

    较陡的线(-x,+ y)与contrast调整为正相关;它的(x,y)坐标是(255 - contrast255 + contrast)。较浅的线(负contrast)的坐标以相同的方式找到。请注意,的最大意义contrast将为255 -最大值(255,255)的初始点可以转换,从而产生一条垂直线(全对比度,全黑或全白)或一条水平线(无对比度,全灰色)。

    所以现在我们在新行上有了两个点的坐标-(0,0)和(255 - contrast255 + contrast)。我们将其插入斜率方程式,然后使用之前的所有部分将其插入全线方程式:



    数学上会注意到结果mfactor是标量(无单位)值;您可以为contrast使用任何想要的范围,只要它与255计算中的常数(factor)相匹配即可。例如,contrast范围为+/-100factor = (100 + contrast)/(100.01 - contrast),这是我真正用来消除缩放到255的步骤;我只是在顶部的代码中保留了255,以简化说明。

    关于“魔术”的注意事项259

    source article使用“魔术” 259,尽管作者承认他不记得为什么:



    259实际上应该是255或256-每个像素的每个通道可能的非零值的数量。请注意,在原始factor计算中,259/255会抵消-从技术上讲是1.01,但是最终值是整数,因此对于所有实际目的都是1。因此,可以舍弃该外部术语。但是,实际上使用255作为分母中的常数会在公式中引入除以零误差的可能性。调整为稍大的值(例如259)可以避免此问题,而不会给结果带来明显的误差。我选择使用255.01来代替,因为错误降低了(希望)对于新手来说似乎不太“神奇”。

    据我所知,it doesn't make much difference which you use-您获得相同的值,除了在低对比度值的窄带中具有较小的对称差异,而正对比度的增加较低。我很好奇要反复往返两个版本并与原始数据进行比较,但是这个答案已经花了太长时间。 :)

    关于html - HTML5 Canvas图像对比度,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/10521978/

    10-11 00:51