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我正在写一个高斯滤镜，我的目标是尽可能地匹配 Photoshop 中的高斯模糊滤镜。这是我的第一次图像处理尝试。我遇到的一些问题/问题是......

用我的滤镜进一步模糊图像会使它变暗，而 Photoshop 似乎会使它变亮。

我使用的偏差值（“sigma”，在我的代码中）是 r/3，这导致矩阵内的高斯曲线接近 0.0001……有没有更好的方法来确定这个值？

Photoshop（或大多数人）如何处理这种模糊的图像边框？

int matrixDimension = (radius*2)+1;
float sigma = radius/3;
float twoSigmaSquared = 2*pow(sigma, 2);
float oneOverSquareRootOfTwoPiSigmaSquared = 1/(sqrt(M_PI*twoSigmaSquared));

float kernel[matrixDimension];

int index = 0;
for (int offset = -radius; offset <= radius; offset++) {

    float xSquared = pow(offset, 2);
    float exponent = -(xSquared/twoSigmaSquared);
    float eToThePower = pow(M_E, exponent);
    float multFactor = oneOverSquareRootOfTwoPiSigmaSquared*eToThePower;

    kernel[index] = multFactor;

    index++;
}

//Normalize the kernel such that all its values will add to 1
float sum = 0;
for (int i = 0; i < matrixDimension; i++) {
    sum += kernel[i];
}
for (int i = 0; i < matrixDimension; i++) {
    kernel[i] = kernel[i]/sum;
}

//Blur horizontally
for (int row = 0; row < imageHeight; row++) {
    for (int column = 0; column < imageWidth; column++) {

        int currentPixel = (row*imageWidth)+column;

        int sum1 = 0;
        int sum2 = 0;
        int sum3 = 0;
        int sum4 = 0;

        int index = 0;
        for (int offset = -radius; offset <= radius; offset++) {
            if (!(column+offset < 0) && !(column+offset > imageWidth-1)) {

                int firstByteOfPixelWereLookingAtInSrcData = (currentPixel+offset)*4;

                int in1 = srcData[firstByteOfPixelWereLookingAtInSrcData];
                int in2 = srcData[firstByteOfPixelWereLookingAtInSrcData+1];
                int in3 = srcData[firstByteOfPixelWereLookingAtInSrcData+2];
                int in4 = srcData[firstByteOfPixelWereLookingAtInSrcData+3];

                sum1 += (int)(in1 * kernel[index]);
                sum2 += (int)(in2 * kernel[index]);
                sum3 += (int)(in3 * kernel[index]);
                sum4 += (int)(in4 * kernel[index]);
            }

            index++;
        }

        int currentPixelInData = currentPixel*4;

        destData[currentPixelInData] = sum1;
        destData[currentPixelInData+1] = sum2;
        destData[currentPixelInData+2] = sum3;
        destData[currentPixelInData+3] = sum4;

    }
}

//Blur vertically
for (int row = 0; row < imageHeight; row++) {
    for (int column = 0; column < imageWidth; column++) {

        int currentPixel = (row*imageWidth)+column;

        int sum1 = 0;
        int sum2 = 0;
        int sum3 = 0;
        int sum4 = 0;

        int index = 0;
        for (int offset = -radius; offset <= radius; offset++) {
            if (!(row+offset < 0) && !(row+offset > imageHeight-1)) {

                int firstByteOfPixelWereLookingAtInSrcData = (currentPixel+(offset*imageWidth))*4;

                int in1 = destData[firstByteOfPixelWereLookingAtInSrcData];
                int in2 = destData[firstByteOfPixelWereLookingAtInSrcData+1];
                int in3 = destData[firstByteOfPixelWereLookingAtInSrcData+2];
                int in4 = destData[firstByteOfPixelWereLookingAtInSrcData+3];

                sum1 += (int)(in1 * kernel[index]);
                sum2 += (int)(in2 * kernel[index]);
                sum3 += (int)(in3 * kernel[index]);
                sum4 += (int)(in4 * kernel[index]);
            }

            index++;
        }

        int currentPixelInData = currentPixel*4;

        finalData[currentPixelInData] = sum1;
        finalData[currentPixelInData+1] = sum2;
        finalData[currentPixelInData+2] = sum3;
        finalData[currentPixelInData+3] = sum4;

    }
}
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2 回答 2

8

要对滤波器进行逆向工程，您需要找到它的脉冲响应。在一个非常暗的值的背景上，比如 32，放置一个近乎白色的像素，比如 223。你不想使用 0 和 255，因为一些过滤器会尝试创建超出起始值的值。在此图像上运行过滤器，并获取输出值并将它们从 0.0 拉伸到 1.0: (value-32)/(223-32)。现在您有了模拟过滤器所需的确切权重。

有很多方法可以处理图像边缘。我建议采用过滤器权重并将它们相加，然后将结果除以该总和；如果您试图超越边缘，请对像素值和该像素上的过滤器权重使用 0.0。

于 2011-03-05T03:45:56.687 回答
1

边界条件有时取决于您正在做什么以及您正在使用什么样的数据，但我认为对于通用图像处理，最好的办法是将边界处的值扩展到图像边缘之外。当然不是字面意思，但是如果过滤器试图读取图像边界之外的像素，则替换图像边缘上最近像素的值。这实际上与将行限制在 0 和高度之间，将列限制在 0 和宽度之间是一样的。

于 2011-03-05T02:44:06.920 回答