java实现图片识别

import javax.imageio.ImageIO;
import java.awt.*;
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.InputStream;

public class ImagePHash {

    public static void main(String[] args) {
        try {
            ImagePHash imagePHash = new ImagePHash();
            String file = "C:\\Users\\1.png";
            String file2 = "C:\\Users\\1.png";
            String pHashString = imagePHash.getHash(new FileInputStream(new File(file)));
            String pHashString2 = imagePHash.getHash(new FileInputStream(new File(file2)));
            // 差异值（0代表两张图片完全一样）
            System.out.println(imagePHash.distance(pHashString, pHashString2));
        } catch (Exception e) {
            e.printStackTrace();
        }
    }
    
    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;
    }


    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. 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);


        /*
         * 5. 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;
    }

    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;
    }


}