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@binzo
Last active February 24, 2023 10:43
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android相似图片算法实现
public class pHash {
/**
* pHash算法流程
* 1.缩小图片,最佳大小为32*32
* 2.转化成灰度图
* 3.转化为DCT图
* 4.取dct图左上角8*8的范围
* 5.计算所有点的平均值
* 6.8*8的范围刚好64个点,计算出64位的图片指纹,如果小于平均值记为0,反之记为1,指纹顺序可以随机,但是每张图片的指纹的顺序应该保持一致
* 7.最后比较两张图片指纹的汉明距离,越小表示越相识
*
*/
//获取指纹,long刚好64位,方便存放
public static long dctImageHash(Bitmap src, boolean recycle) throws IOException {
//由于计算dct需要图片长宽相等,所以统一取32
int length = 32;
//缩放图片
Bitmap bitmap = scaleBitmap(src, recycle, length);
//获取灰度图
int[] pixels = createGrayImage(bitmap, length);
//先获得32*32的dct,再取dct左上角8*8的区域
return computeHash(DCT8(pixels, length));
}
private static int[] createGrayImage(Bitmap src, int length) {
int[] pixels = new int[length * length];
src.getPixels(pixels, 0, length, 0, 0, length, length);
src.recycle();
for (int i = 0; i < pixels.length; i++) {
int gray = computeGray(pixels[i]);
pixels[i] = Color.rgb(gray, gray, gray);
}
return pixels;
}
//缩放成宽高一样的图片
private static Bitmap scaleBitmap(Bitmap src, boolean recycle, float length) throws IOException {
if (src == null) {
throw new IOException("invalid image");
}
int width = src.getWidth();
int height = src.getHeight();
if (width == 0 || height == 0) {
throw new IOException("invalid image");
}
Matrix matrix = new Matrix();
matrix.postScale(length / width, length / height);
Bitmap bitmap = Bitmap.createBitmap(src, 0, 0, width, height, matrix, false);
if (recycle) {
src.recycle();
}
return bitmap;
}
//计算hash值
private static long computeHash(double[] pxs) {
double t = 0;
for (double i : pxs) {
t += i;
}
double median = t / pxs.length;
long one = 0x0000000000000001;
long hash = 0x0000000000000000;
for (double current : pxs) {
if (current > median)
hash |= one;
one = one << 1;
}
return hash;
}
/**
*计算灰度值
* 计算公式Gray = R*0.299 + G*0.587 + B*0.114
* 由于浮点数运算性能较低,转换成位移运算
* 向右每位移一位,相当于除以2
*
*/
private static int computeGray(int pixel) {
int red = Color.red(pixel);
int green = Color.green(pixel);
int blue = Color.blue(pixel);
return (red * 38 + green * 75 + blue * 15) >> 7;
}
//取dct图左上角8*8的区域
private static double[] DCT8(int[] pix, int n) {
double[][] iMatrix = DCT(pix, n);
double px[] = new double[8 * 8];
for (int i = 0; i < 8; i++) {
System.arraycopy(iMatrix[i], 0, px, i * 8, 8);
}
return px;
}
/**
* 离散余弦变换
*
* 计算公式为:系数矩阵*图片矩阵*转置系数矩阵
*
* @param pix 原图像的数据矩阵
* @param n 原图像(n*n)
* @return 变换后的矩阵数组
*/
private static double[][] DCT(int[] pix, int n) {
double[][] iMatrix = new double[n][n];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
iMatrix[i][j] = (double) (pix[i * n + j]);
}
}
double[][] quotient = coefficient(n); //求系数矩阵
double[][] quotientT = transposingMatrix(quotient, n); //转置系数矩阵
double[][] temp;
temp = matrixMultiply(quotient, iMatrix, n);
iMatrix = matrixMultiply(temp, quotientT, n);
return iMatrix;
}
/**
* 矩阵转置
*
* @param matrix 原矩阵
* @param n 矩阵(n*n)
* @return 转置后的矩阵
*/
private static double[][] transposingMatrix(double[][] matrix, int n) {
double nMatrix[][] = new double[n][n];
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
nMatrix[i][j] = matrix[j][i];
}
}
return nMatrix;
}
/**
* 求离散余弦变换的系数矩阵
*
* @param n n*n矩阵的大小
* @return 系数矩阵
*/
private static double[][] coefficient(int n) {
double[][] coeff = new double[n][n];
double sqrt = Math.sqrt(1.0 / n);
double sqrt1 = Math.sqrt(2.0 / n);
for (int i = 0; i < n; i++) {
coeff[0][i] = sqrt;
}
for (int i = 1; i < n; i++) {
for (int j = 0; j < n; j++) {
coeff[i][j] = sqrt1 * Math.cos(i * Math.PI * (j + 0.5) / n);
}
}
return coeff;
}
/**
* 矩阵相乘
*
* @param A 矩阵A
* @param B 矩阵B
* @param n 矩阵的大小n*n
* @return 结果矩阵
*/
private static double[][] matrixMultiply(double[][] A, double[][] B, int n) {
double nMatrix[][] = new double[n][n];
double t;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
t = 0;
for (int k = 0; k < n; k++) {
t += A[i][k] * B[k][j];
}
nMatrix[i][j] = t;
}
}
return nMatrix;
}
/**
* 计算两个图片指纹的汉明距离
*
* @param hash1 指纹1
* @param hash2 指纹2
* @return 返回汉明距离 也就是64位long型不相同的位的个数
*/
public static int hammingDistance(long hash1, long hash2) {
long x = hash1 ^ hash2;
final long m1 = 0x5555555555555555L;
final long m2 = 0x3333333333333333L;
final long h01 = 0x0101010101010101L;
final long m4 = 0x0f0f0f0f0f0f0f0fL;
x -= (x >> 1) & m1;
x = (x & m2) + ((x >> 2) & m2);
x = (x + (x >> 4)) & m4;
return (int) ((x * h01) >> 56);
}
}
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