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java获取两张图片的相似度
阅读量:7209 次
发布时间:2019-06-29

本文共 6269 字,大约阅读时间需要 20 分钟。

hot3.png

package com.sinosoft.lis.utils;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.IOException;import javax.imageio.ImageIO;/** * 图片相似性 */public class ImageSimilarity {   public static int size = 32;   public static int smallerSize = 8;   // DCT function stolen from   // http://stackoverflow.com/questions/4240490/problems-with-dct-and-idct-algorithm-in-java   private static double[] c;   static {      c = new double[size];      for (int i = 1; i < size; i++) {         c[i] = 1;      }      c[0] = 1 / Math.sqrt(2.0);   }   /**    * 通过汉明距离计算相似度    *     * @param hash1    * @param hash2    * @return    */   public static double calSimilarity(String hash1, String hash2) {      return calSimilarity(getHammingDistance(hash1, hash2));   }   /**    * 通过汉明距离计算相似度    *     * @param hammingDistance    * @return    */   public static double calSimilarity(int hammingDistance) {      int length = size * size;      double similarity = (length - hammingDistance) / (double) length;      // 使用指数曲线调整相似度结果      similarity = Math.pow(similarity, 2);      return similarity;   }   /**    * 通过汉明距离计算相似度    *     * @param image1    * @param image2    * @return    * @throws IOException    */   public static double calSimilarity(File image1, File image2) throws IOException {      return calSimilarity(getHammingDistance(image1, image2));   }   /**    * 获得汉明距离    *     * @param hash1    * @param hash2    * @return    */   public static int getHammingDistance(String hash1, String hash2) {      int counter = 0;      for (int k = 0; k < hash1.length(); k++) {         if (hash1.charAt(k) != hash2.charAt(k)) {            counter++;         }      }      return counter;   }   /**    * 获得汉明距离    *     * @param image1    * @param image2    * @return    * @throws IOException    */   public static int getHammingDistance(File image1, File image2) throws IOException {      return getHammingDistance(getHash(image1), getHash(image2));   }   /**    * 返回二进制字符串,类似“001010111011100010”,可用于计算汉明距离    *     * @param imageFile    * @return    * @throws IOException    * @throws Exception    */   public static String getHash(File imageFile) throws IOException {      BufferedImage img = ImageIO.read(imageFile);      /*       * 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);      /*       * 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.       */      StringBuilder hash = new StringBuilder();      for (int x = 0; x < smallerSize; x++) {         for (int y = 0; y < smallerSize; y++) {            if (x != 0 && y != 0) {               hash.append((dctVals[x][y] > avg ? "1" : "0"));            }         }      }      return hash.toString();   }   private static 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 static BufferedImage grayscale(BufferedImage img) {      new ColorConvertOp(ColorSpace.getInstance(ColorSpace.CS_GRAY), null).filter(img, img);      return img;   }   private static int getBlue(BufferedImage img, int x, int y) {      return (img.getRGB(x, y)) & 0xff;   }   private static 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;   }}

调用对比功能,传入两个图片的file对象,图片越相似,数值越接近于1,当图片相同时,等于1

ImageSimilarity.calSimilarity(imageFile1, imageFile2) == 1.0

 

转载于:https://my.oschina.net/u/3768722/blog/2248000

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