java獲取兩張圖片的類似度

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;

/**
 * 圖片類似性
 */
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);
   }

   /**
    * 經過漢明距離計算類似度
    * 
    * @param hash1
    * @param hash2
    * @return
    */
   public static double calSimilarity(String hash1, String hash2) {
      return calSimilarity(getHammingDistance(hash1, hash2));
   }

   /**
    * 經過漢明距離計算類似度
    * 
    * @param hammingDistance
    * @return
    */
   public static double calSimilarity(int hammingDistance) {
      int length = size * size;
      double similarity = (length - hammingDistance) / (double) length;

      // 使用指數曲線調整類似度結果
      similarity = Math.pow(similarity, 2);
      return similarity;
   }

   /**
    * 經過漢明距離計算類似度
    * 
    * @param image1
    * @param image2
    * @return
    * @throws IOException
    */
   public static double calSimilarity(File image1, File image2) throws IOException {
      return calSimilarity(getHammingDistance(image1, image2));
   }

   /**
    * 得到漢明距離
    * 
    * @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;
   }

   /**
    * 得到漢明距離
    * 
    * @param image1
    * @param image2
    * @return
    * @throws IOException
    */
   public static int getHammingDistance(File image1, File image2) throws IOException {
      return getHammingDistance(getHash(image1), getHash(image2));
   }

   /**
    * 返回二進制字符串，相似「001010111011100010」，可用於計算漢明距離
    * 
    * @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;
   }
}

調用對比功能，傳入兩個圖片的file對象，圖片越類似，數值越接近於1，當圖片相同時，等於1

ImageSimilarity.calSimilarity(imageFile1, imageFile2) == 1.0