用汉明距离进行图片相似度检测的Java实现

jopen 12年前

Google、Baidu 等搜索引擎相继推出了以图搜图的功能,测试了下效果还不错~ 那这种技术的原理是什么呢?计算机怎么知道两张图片相似呢?

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。

用汉明距离进行图片相似度检测的Java实现 用汉明距离进行图片相似度检测的Java实现

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。

用汉明距离进行图片相似度检测的Java实现 = 用汉明距离进行图片相似度检测的Java实现 = 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:

 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:

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