码迷,mamicode.com
首页 > 编程语言 > 详细

用汉明距离进行图片相似度检测的Java实现

时间:2015-03-13 15:48:17      阅读:880      评论:0      收藏:0      [点我收藏+]

标签:

根据Neal Krawetz博士的解释,原理非常简单易懂。我们可以用一个快速算法,就达到基本的效果。

这里的关键技术叫做"感知哈希算法"(Perceptual hash algorithm),它的作用是对每张图片生成一个"指纹"(fingerprint)字符串,然后比较不同图片的指纹。结果越接近,就说明图片越相似。

下面是一个最简单的实现:

第一步,缩小尺寸。

将图片缩小到8x8的尺寸,总共64个像素。这一步的作用是去除图片的细节,只保留结构、明暗等基本信息,摒弃不同尺寸、比例带来的图片差异。

技术分享 技术分享

第二步,简化色彩。

将缩小后的图片,转为64级灰度。也就是说,所有像素点总共只有64种颜色。

第三步,计算平均值。

计算所有64个像素的灰度平均值。

第四步,比较像素的灰度。

将每个像素的灰度,与平均值进行比较。大于或等于平均值,记为1;小于平均值,记为0。

第五步,计算哈希值。

将上一步的比较结果,组合在一起,就构成了一个64位的整数,这就是这张图片的指纹。组合的次序并不重要,只要保证所有图片都采用同样次序就行了。

技术分享 = 技术分享 = 8f373714acfcf4d0

得到指纹以后,就可以对比不同的图片,看看64位中有多少位是不一样的。在理论上,这等同于计算"汉明距离"(Hamming distance)。如果不相同的数据位不超过5,就说明两张图片很相似;如果大于10,就说明这是两张不同的图片。

具体的代码实现,可以参见Wote用python语言写的imgHash.py。代码很短,只有53行。使用的时候,第一个参数是基准图片,第二个参数是用来比较的其他图片所在的目录,返回结果是两张图片之间不相同的数据位数量(汉明距离)。

这种算法的优点是简单快速,不受图片大小缩放的影响,缺点是图片的内容不能变更。如果在图片上加几个文字,它就认不出来了。所以,它的最佳用途是根据缩略图,找出原图。

实际应用中,往往采用更强大的pHash算法和SIFT算法,它们能够识别图片的变形。只要变形程度不超过25%,它们就能匹配原图。这些算法虽然更复杂,但是原理与上面的简便算法是一样的,就是先将图片转化成Hash字符串,然后再进行比较。

下面我们来看下上述理论用java来做一个DEMO版的具体实现:

 

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
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 javax.imageio.ImageIO;
/*
* pHash-like image hash.
* Author: Elliot Shepherd (elliot@jarofworms.com
* Based On: http://www.hackerfactor.com/blog/index.php?/archives/432-Looks-Like-It.html
*/
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. 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);
       System.out.println("DCT: " + (System.currentTimeMillis() - start));
        
       /* 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.
        */
       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, 00, 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;
   }
 
   public static void main(String[] args) {
        
       ImagePHash p = new ImagePHash();
       String image1;
       String image2;
       try {
           image1 = p.getHash(new FileInputStream(new File("C:/Users/june/Desktop/1.jpg")));
           image2 = p.getHash(new FileInputStream(new File("C:/Users/june/Desktop/1.jpg")));
           System.out.println("1:1 Score is " + p.distance(image1, image2));
           image1 = p.getHash(new FileInputStream(new File("C:/Users/june/Desktop/1.jpg")));
           image2 = p.getHash(new FileInputStream(new File("C:/Users/june/Desktop/2.jpg")));
           System.out.println("1:2 Score is " + p.distance(image1, image2));
           image1 = p.getHash(new FileInputStream(new File("C:/Users/june/Desktop/1.jpg")));
           image2 = p.getHash(new FileInputStream(new File("C:/Users/june/Desktop/3.jpg")));
           System.out.println("1:3 Score is " + p.distance(image1, image2));
           image1 = p.getHash(new FileInputStream(new File("C:/Users/june/Desktop/2.jpg")));
           image2 = p.getHash(new FileInputStream(new File("C:/Users/june/Desktop/3.jpg")));
           System.out.println("2:3 Score is " + p.distance(image1, image2));
            
           image1 = p.getHash(new FileInputStream(new File("C:/Users/june/Desktop/4.jpg")));
           image2 = p.getHash(new FileInputStream(new File("C:/Users/june/Desktop/5.jpg")));
           System.out.println("4:5 Score is " + p.distance(image1, image2));
            
       catch (FileNotFoundException e) {
           e.printStackTrace();
       catch (Exception e) {
           e.printStackTrace();
       }
 
   }
}

 

 

运行结果为:

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
DCT: 163
DCT: 158
1:1 Score is 0
DCT: 168
DCT: 164
1:2 Score is 4
DCT: 156
DCT: 156
1:3 Score is 3
DCT: 157
DCT: 157
2:3 Score is 1
DCT: 157
DCT: 156
4:5 Score is 21

说明:其中1,2,3是3张非常相似的图片,图片分别加了不同的文字水印,肉眼分辨的不是太清楚,下面会有附图,4、5是两张差异很大的图,图你可以随便找来测试,这两张我就不上传了。

 

结果说明:汉明距离越大表明图片差异越大,如果不相同的数据位不超过5,就说明两张图片很相似;如果大于10,就说明这是两张不同的图片。从结果可以看到1、2、3是相似图片,4、5差异太大,是两张不同的图片。

附:图1、2、3

技术分享

图1

技术分享

图2

技术分享

图3

参考地址:

代码参考:http://pastebin.com/Pj9d8jt5
原理参考:http://www.ruanyifeng.com/blog/2011/07/principle_of_similar_image_search.html
汉明距离:http://baike.baidu.com/view/725269.htm

来自:http://stackoverflow.com/questions/6971966/how-to-measure-percentage-similarity-between-two-images

用汉明距离进行图片相似度检测的Java实现

标签:

原文地址:http://www.cnblogs.com/lixun/p/4334977.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!