HTML5 Canvas图像对比度
Sch*_*rgh 22 html5 image-processing contrast html5-canvas
我一直在编写一个图像处理程序,它通过HTML5画布像素处理来应用效果.我已经实现了Thresholding,Vintaging和ColorGradient像素操作,但令人难以置信的是我无法改变图像的对比度!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并且对比效果较少而且我不打算使用任何Javascript库,因为我试图在本地实现这些效果.
基本像素操作代码:
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];
}
像素操作示例
值处于RGB模式,这意味着data [i]是红色.所以如果data [i] = data [i]*2; 对于该像素的红色通道,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:我不是要求你们完成代码!那只会是一个忙!我要求一个算法(甚至伪代码),它显示像素操作中的对比度是如何可能的! 如果有人能够在HTML5画布中为图像对比提供一个好的算法,我会很高兴.
Bri*_*ian 26
在尝试了Schahriar SaffarShargh的答案之后,它的表现并不像对比应该表现的那样.我终于遇到了这个算法,它就像一个魅力!
有关该算法的其他信息,请阅读本文及其注释部分.
function contrastImage(imageData, contrast) {
var data = imageData.data;
var factor = (259 * (contrast + 255)) / (255 * (259 - contrast));
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;
}
用法:
var newImageData = contrastImage(imageData, 30);
希望这对某人节省时间.干杯!
bri*_*ins 19
更快的选择(基于Escher的方法)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; 这个版本在数学上是相同的,但运行速度要快得多.
原始答案
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一系列+/- 100而不是原始系列+/- 255.对于不了解底层概念的用户或程序员来说,百分比值似乎更直观.此外,我的用法总是与UI控件绑定; 从-100%到+ 100%的范围允许我直接标记和绑定控制值,而不是调整或解释它.此算法不包括范围检查,即使计算的值可能远远超出允许范围 - 这是因为ImageData对象下面的数组是a
Uint8ClampedArray.正如MSDN解释的那样,Uint8ClampedArray为您处理范围检查:
"如果您指定的值超出[0,255]的范围,则会设置0或255."
用法
Note that while the underlying formula is fairly symmetric (allows round-tripping), data is lost at high levels of filtering because pixels only allow integer values. For example, by the time you de-saturate an image to extreme levels (>95% or so), all the pixels are basically a uniform medium gray (within a few digits of the average possible value of 128). Turning the contrast back up again results in a flattened color range.
Also, order of operations is important when applying multiple contrast adjustments - saturated values "blow out" (exceed the clamped max value of 255) quickly, meaning highly saturating and then de-saturating will result in a darker image overall. De-saturating and then saturating however doesn't have as much data loss, because the highlight and shadow values get muted, instead of clipped (see explanation below).
Generally speaking, when applying multiple filters it's better to start each operation with the original data and re-apply each adjustment in turn, rather than trying to reverse a previous change - at least for image quality. Performance speed or other demands may dictate differently for each situation.
Code Example:
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 = "data:image/png;base64,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div {text-align:center; width:120px; float:left}<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
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Original
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Round-trip <br/> (-98%, +98%)
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</div>Explanation
(Disclaimer - I am not an image specialist or a mathematician. I am trying to provide a common-sense explanation with minimal technical details. Some hand-waving below, e.g. 255=256 to avoid indexing issues, and 127.5=128, for simplifying the numbers.)
Since, for a given pixel, the possible number of non-zero values for a color channel is 255, the "no-contrast", average value of a pixel is 128 (or 127, or 127.5 if you want argue, but the difference is negligible). For purposed of this explanation, the amount of "contrast" is the distance from the current value to the average value (128). Adjusting the contrast means increasing or decreasing the difference between the current value and the average value.
算法解决的问题是:
- 选择一个恒定因子来调整对比度
- 对于每个像素的每个颜色通道,通过该常数因子缩放"对比度"(与平均值的距离)
或者,正如CSS规范中暗示的那样,只需选择一条线的斜率和截距:
<feFuncR type="linear" slope="[amount]" intercept="-(0.5 * [amount]) + 0.5"/>
注意这个词type='linear'; 我们正在RGB色彩空间中进行线性对比度调整,而不是二次缩放功能,基于亮度的调整或直方图匹配.
If you recall from geometry class, the formula for a line is y=mx+b. y is the final value we are after, the slope m is the contrast (or factor), x is the initial pixel value, and b is the intercept of the y-axis (x=0), which shifts the line vertically. Recall also that since the y-intercept is not at the origin (0,0), the formula can also be represented as y=m(x-a)+b, where a is the x-offset shifting the line horizontally.
For our purposes, this graph represents the input value (x-axis) and the result (y-axis). We already know that b, the y-intercept (for m=0, no contrast) must be 128 (which we can check against the 0.5 from the spec - 0.5*the full range of 256 = 128). x is our original value, so all we need is to figure out the slope m and x-offset a.
First, the slope m is "rise over run", or (y2-y1)/(x2-x1) - so we need 2 points known to be on the desired line. Finding these points requires bringing a few things together:
- Our function takes the shape of a line-intercept graph
- The y-intercept is at
b = 128- regardless of the slope (contrast). - The maximum expected 'y' value is 255, and the minimum is 0
- The range of possible 'x' values is 256
- A neutral value should always stay neutral: 128 => 128 regardless of slope
- A contrast adjustment of
0should result in no change between input and output; that is, a 1:1 slope.
Taking all these together, we can deduce that regardless of the contrast (slope) applied, our resulting line will be centered at (and pivot around) 128,128. Since our y-intercept is non-zero, the x-intercept is also non-zero; we know the x-range is 256 wide and is centered in the middle, so it must be offset by half of the possible range: 256/2 = 128.
So now for y=m(x-a)+b, we know everything except m. Recall two more important points from geometry class:
- Lines have the same slope even if their location changes; that is,
mstays the same regardless of the values ofaandb. - The slope of a line can be found using any 2 points on the line
To simplify the slope discussion, let's move the coordinate origin to the x-intercept (-128) and ignore a and b for a moment. Our original line will now pivot through (0,0), and we know a second point on the line lies away the full range of both x (input) and y (output) at (255,255).
We'll let the new line pivot at (0,0), so we can use that as one of the points on the new line that will follow our final contrast slope m. The second point can be determined by moving the current end at (255,255) by some amount; since we are limited to a single input (contrast) and using a linear function, this second point will be moved equally in the x and y directions on our graph.
The (x,y) coordinates of the 4 possible new points will be 255 +/- contrast. Since increasing or decreasing both x and y would keep us on the original 1:1 line, let's just look at +x, -y and -x, +y as shown.
The steeper line (-x, +y) is associated with a positive contrast adjustment; it's (x,y) coordinates are (255 - contrast,255 + contrast). The coordinates of the shallower line (negative contrast) are found the same way. Notice that the biggest meaningful value of contrast will be 255 - the most that the initial point of (255,255) can be translated before resulting in a vertical line (full contrast, all black or white) or a horizontal line (no contrast, all gray).
So now we have the coordinates of two points on our new line - (0,0) and (255 - contrast,255 + contrast). We plug this into the slope equation, and then plug that into the full line equation, using all the parts from before:
y = m(x-a) + b
m=(y2-y1)/(x2-x1)=>
((255 + contrast) - 0)/((255 - contrast) - 0)=>
(255 + contrast)/(255 - contrast)
a = 128
b = 128
y = (255 + contrast)/(255 - contrast) * (x - 128) + 128QED
The math-minded will notice that the resulting m or factor is a scalar (unitless) value; you can use any range you want for contrast as long as it matches the constant (255) in the factor calculation. For example, a contrast range of +/-100 and factor = (100 + contrast)/(100.01 - contrast), which is was I really use to eliminate the step of scaling to 255; I just left 255 in the code at the top to simplify the explanation.
Note about the "magic" 259
The source article uses a "magic" 259, although the author admits he doesn't remember why:
"I can’t remember if I had calculated this myself or if I’ve read it in a book or online.".
259 should really be 255 or perhaps 256 - the number of possible non-zero values for each channel of each pixel. Note that in the original factor calculation, 259/255 cancels out - technically 1.01, but final values are whole integers so 1 for all practical purposes. So this outer term can be discarded. Actually using 255 for the constant in the denominator, though, introduces the possibility of a Divide By Zero error in the formula; adjusting to a slightly larger value (say, 259) avoids this issue without introducing significant error to the results. I chose to use 255.01 instead as the error is lower and it (hopefully) seems less "magic" to a newcomer.
As far as I can tell though, it doesn't make much difference which you use - you get identical values except for minor, symmetric differences in a narrow band of low contrast values with a low positive contrast increase. I'd be curious to round-trip both versions repeatedly and compare to the original data, but this answer already took way too long. :)