java实现图片识别

package IMGPACK;






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 java.sql.Connection;
import java.sql.PreparedStatement;
import java.sql.ResultSet;
import java.util.concurrent.ExecutionException;
import java.util.concurrent.FutureTask;


import javax.imageio.ImageIO;


import com.sun.corba.se.impl.orbutil.closure.Future;


import jdbc.JDBC;


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.缩小尺寸 pHash以小图片开始，但图片大于8*8，32*32是最好的。这样做的目的是简化了DCT的计算，而不是减小频率。
*/
img = resize(img, size, size);


/*
* 2. 简化色彩 将图片转化成灰度图像，进一步简化计算量。
*/
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.计算DCT DCT是把图片分解频率聚集和梯状形，虽然JPEG使用8*8的DCT变换，在这里使用32*32的DCT变换。
*/
double[][] dctVals = applyDCT(vals);


/*
* 4.缩小DCT 虽然DCT的结果是32*32大小的矩阵，但我们只要保留左上角的8*8的矩阵，这部分呈现了图片中的最低频率。
*/
/*
* 5. 计算平均值 如同均值哈希一样，计算DCT的均值，
*/
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. 进一步减小DCT 这是最主要的一步，根据8*8的DCT矩阵， 设置0或1的64位的hash值，大于
* 等于DCT均值的设为”1”，小于DCT均值的设为“0”。结果并不能告诉我们真
* 实性的低频率，只能粗略地告诉我们相对于平均值频率的相对比例。只要图 片的整体结构保持不变，hash结果值就不变。能够避免伽马校正或颜色直方
* 图被调整带来的影响。
*/
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;
}


}