用汉明距离进行图片相似度检测的Java实现

根据Neal Krawetz博士的解释，原理非常简单易懂。我们可以用一个快速算法，就达到基本的效果。

这里的关键技术叫做"感知哈希算法"（Perceptual hash algorithm），它的作用是对每张图片生成一个"指纹"（fingerprint）字符串，然后比较不同图片的指纹。结果越接近，就说明图片越相似。

下面是一个最简单的实现：

第一步，缩小尺寸。

将图片缩小到8x8的尺寸，总共64个像素。这一步的作用是去除图片的细节，只保留结构、明暗等基本信息，摒弃不同尺寸、比例带来的图片差异。

 

第二步，简化色彩。

将缩小后的图片，转为64级灰度。也就是说，所有像素点总共只有64种颜色。

第三步，计算平均值。

计算所有64个像素的灰度平均值。

第四步，比较像素的灰度。

将每个像素的灰度，与平均值进行比较。大于或等于平均值，记为1；小于平均值，记为0。

第五步，计算哈希值。

将上一步的比较结果，组合在一起，就构成了一个64位的整数，这就是这张图片的指纹。组合的次序并不重要，只要保证所有图片都采用同样次序就行了。

 =  = 8f373714acfcf4d0

得到指纹以后，就可以对比不同的图片，看看64位中有多少位是不一样的。在理论上，这等同于计算"汉明距离"（Hamming distance）。如果不相同的数据位不超过5，就说明两张图片很相似；如果大于10，就说明这是两张不同的图片。

具体的代码实现，可以参见Wote用python语言写的imgHash.py。代码很短，只有53行。使用的时候，第一个参数是基准图片，第二个参数是用来比较的其他图片所在的目录，返回结果是两张图片之间不相同的数据位数量（汉明距离）。

这种算法的优点是简单快速，不受图片大小缩放的影响，缺点是图片的内容不能变更。如果在图片上加几个文字，它就认不出来了。所以，它的最佳用途是根据缩略图，找出原图。

实际应用中，往往采用更强大的pHash算法和SIFT算法，它们能够识别图片的变形。只要变形程度不超过25%，它们就能匹配原图。这些算法虽然更复杂，但是原理与上面的简便算法是一样的，就是先将图片转化成Hash字符串，然后再进行比较。

下面我们来看下上述理论用java来做一个DEMO版的具体实现：

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import java.awt.Graphics2D; import java.awt.color.ColorSpace; import java.awt.image.BufferedImage; import java.awt.image.ColorConvertOp; import java.io.File; import java.io.FileInputStream; import java.io.FileNotFoundException; import java.io.InputStream;   import javax.imageio.ImageIO; /* * pHash-like image hash. * Author: Elliot Shepherd (elliot@jarofworms.com * Based On: http://www.hackerfactor.com/blog/index.php?/archives/432-Looks-Like-It.html */ public class ImagePHash {      private int size = 32;    private int smallerSize = 8;         public ImagePHash() {        initCoefficients();    }         public ImagePHash(int size, int smallerSize) {        this.size = size;        this.smallerSize = smallerSize;                 initCoefficients();    }         public int distance(String s1, String s2) {        int counter = 0;        for (int k = 0; k < s1.length();k++) {            if(s1.charAt(k) != s2.charAt(k)) {                counter++;            }        }        return counter;    }         // Returns a 'binary string' (like. 001010111011100010) which is easy to do a hamming distance on.    public String getHash(InputStream is) throws Exception {        BufferedImage img = ImageIO.read(is);                 /* 1. Reduce size.         * Like Average Hash, pHash starts with a small image.         * However, the image is larger than 8x8; 32x32 is a good size.         * This is really done to simplify the DCT computation and not         * because it is needed to reduce the high frequencies.         */        img = resize(img, size, size);                 /* 2. Reduce color.         * The image is reduced to a grayscale just to further simplify         * the number of computations.         */        img = grayscale(img);                 double[][] vals = new double[size][size];                 for (int x = 0; x < img.getWidth(); x++) {            for (int y = 0; y < img.getHeight(); y++) {                vals[x][y] = getBlue(img, x, y);            }        }                 /* 3. Compute the DCT.         * The DCT separates the image into a collection of frequencies         * and scalars. While JPEG uses an 8x8 DCT, this algorithm uses         * a 32x32 DCT.         */        long start = System.currentTimeMillis();        double[][] dctVals = applyDCT(vals);        System.out.println("DCT: " + (System.currentTimeMillis() - start));                 /* 4. Reduce the DCT.         * This is the magic step. While the DCT is 32x32, just keep the         * top-left 8x8. Those represent the lowest frequencies in the         * picture.         */        /* 5. Compute the average value.         * Like the Average Hash, compute the mean DCT value (using only         * the 8x8 DCT low-frequency values and excluding the first term         * since the DC coefficient can be significantly different from         * the other values and will throw off the average).         */        double total = 0;                 for (int x = 0; x < smallerSize; x++) {            for (int y = 0; y < smallerSize; y++) {                total += dctVals[x][y];            }        }        total -= dctVals[0][0];                 double avg = total / (double) ((smallerSize * smallerSize) - 1);             /* 6. Further reduce the DCT.         * This is the magic step. Set the 64 hash bits to 0 or 1         * depending on whether each of the 64 DCT values is above or         * below the average value. The result doesn't tell us the         * actual low frequencies; it just tells us the very-rough         * relative scale of the frequencies to the mean. The result         * will not vary as long as the overall structure of the image         * remains the same; this can survive gamma and color histogram         * adjustments without a problem.         */        String hash = "";                 for (int x = 0; x < smallerSize; x++) {            for (int y = 0; y < smallerSize; y++) {                if (x != 0 && y != 0) {                    hash += (dctVals[x][y] > avg?"1":"0");                }            }        }                 return hash;    }         private BufferedImage resize(BufferedImage image, int width,    int height) {        BufferedImage resizedImage = new BufferedImage(width, height, BufferedImage.TYPE_INT_ARGB);        Graphics2D g = resizedImage.createGraphics();        g.drawImage(image, 0, 0, width, height, null);        g.dispose();        return resizedImage;    }         private ColorConvertOp colorConvert = new ColorConvertOp(ColorSpace.getInstance(ColorSpace.CS_GRAY), null);      private BufferedImage grayscale(BufferedImage img) {        colorConvert.filter(img, img);        return img;    }         private static int getBlue(BufferedImage img, int x, int y) {        return (img.getRGB(x, y)) & 0xff;    }         // DCT function stolen from http://stackoverflow.com/questions/4240490/problems-with-dct-and-idct-algorithm-in-java      private double[] c;    private void initCoefficients() {        c = new double[size];                 for (int i=1;i<size;i++) {            c[i]=1;        }        c[0]=1/Math.sqrt(2.0);    }         private double[][] applyDCT(double[][] f) {        int N = size;                 double[][] F = new double[N][N];        for (int u=0;u<N;u++) {          for (int v=0;v<N;v++) {            double sum = 0.0;            for (int i=0;i<N;i++) {              for (int j=0;j<N;j++) {                sum+=Math.cos(((2*i+1)/(2.0*N))*u*Math.PI)*Math.cos(((2*j+1)/(2.0*N))*v*Math.PI)*(f[i][j]);              }            }            sum*=((c[u]*c[v])/4.0);            F[u][v] = sum;          }        }        return F;    }      public static void main(String[] args) {                 ImagePHash p = new ImagePHash();        String image1;        String image2;        try {            image1 = p.getHash(new FileInputStream(new File("C:/Users/june/Desktop/1.jpg")));            image2 = p.getHash(new FileInputStream(new File("C:/Users/june/Desktop/1.jpg")));            System.out.println("1:1 Score is " + p.distance(image1, image2));            image1 = p.getHash(new FileInputStream(new File("C:/Users/june/Desktop/1.jpg")));            image2 = p.getHash(new FileInputStream(new File("C:/Users/june/Desktop/2.jpg")));            System.out.println("1:2 Score is " + p.distance(image1, image2));            image1 = p.getHash(new FileInputStream(new File("C:/Users/june/Desktop/1.jpg")));            image2 = p.getHash(new FileInputStream(new File("C:/Users/june/Desktop/3.jpg")));            System.out.println("1:3 Score is " + p.distance(image1, image2));            image1 = p.getHash(new FileInputStream(new File("C:/Users/june/Desktop/2.jpg")));            image2 = p.getHash(new FileInputStream(new File("C:/Users/june/Desktop/3.jpg")));            System.out.println("2:3 Score is " + p.distance(image1, image2));                         image1 = p.getHash(new FileInputStream(new File("C:/Users/june/Desktop/4.jpg")));            image2 = p.getHash(new FileInputStream(new File("C:/Users/june/Desktop/5.jpg")));            System.out.println("4:5 Score is " + p.distance(image1, image2));                     } catch (FileNotFoundException e) {            e.printStackTrace();        } catch (Exception e) {            e.printStackTrace();        }      } }

运行结果为：

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DCT: 163 DCT: 158 1:1 Score is 0 DCT: 168 DCT: 164 1:2 Score is 4 DCT: 156 DCT: 156 1:3 Score is 3 DCT: 157 DCT: 157 2:3 Score is 1 DCT: 157 DCT: 156 4:5 Score is 21

说明：其中1,2,3是3张非常相似的图片，图片分别加了不同的文字水印，肉眼分辨的不是太清楚，下面会有附图，4、5是两张差异很大的图，图你可以随便找来测试，这两张我就不上传了。

结果说明：汉明距离越大表明图片差异越大，如果不相同的数据位不超过5，就说明两张图片很相似；如果大于10，就说明这是两张不同的图片。从结果可以看到1、2、3是相似图片，4、5差异太大，是两张不同的图片。

附：图1、2、3

图1

图2

图3

参考地址：

代码参考：http://pastebin.com/Pj9d8jt5
原理参考：http://www.ruanyifeng.com/blog/2011/07/principle_of_similar_image_search.html
汉明距离：http://baike.baidu.com/view/725269.htm

来自：http://stackoverflow.com/questions/6971966/how-to-measure-percentage-similarity-between-two-images