I believe that dropping the size of the image down to an almost icon size, say 48x48, then converting to greyscale, then taking the difference between pixels, or Delta, should work well. Because we're comparing the change in pixel color, rather than the actual pixel color, it won't matter if the image is slightly lighter or darker. Large changes will matter since pixels getting too light/dark will be lost. You can apply this across one row, or as many as you like to increase the accuracy. At most you'd have 47x47=2,209 subtractions to make in order to form a comparable Key.

如果您有大量的图像，请查看Bloom过滤器，它使用多个散列来获得概率高但效率高的结果。如果图像的数量不是很大，那么像md5这样的加密散列应该足够了。

下面是解决这个问题的三种方法(还有很多其他方法)。

The first is a standard approach in computer vision, keypoint matching. This may require some background knowledge to implement, and can be slow. The second method uses only elementary image processing, and is potentially faster than the first approach, and is straightforward to implement. However, what it gains in understandability, it lacks in robustness -- matching fails on scaled, rotated, or discolored images. The third method is both fast and robust, but is potentially the hardest to implement.

关键点匹配

Better than picking 100 random points is picking 100 important points. Certain parts of an image have more information than others (particularly at edges and corners), and these are the ones you'll want to use for smart image matching. Google "keypoint extraction" and "keypoint matching" and you'll find quite a few academic papers on the subject. These days, SIFT keypoints are arguably the most popular, since they can match images under different scales, rotations, and lighting. Some SIFT implementations can be found here.

关键点匹配的一个缺点是简单实现的运行时间:O(n^2m)，其中n是每张图像中的关键点数量，m是数据库中的图像数量。一些聪明的算法可能会更快地找到最接近的匹配，比如四叉树或二进制空间分区。

备选方案:直方图法

Another less robust but potentially faster solution is to build feature histograms for each image, and choose the image with the histogram closest to the input image's histogram. I implemented this as an undergrad, and we used 3 color histograms (red, green, and blue), and two texture histograms, direction and scale. I'll give the details below, but I should note that this only worked well for matching images VERY similar to the database images. Re-scaled, rotated, or discolored images can fail with this method, but small changes like cropping won't break the algorithm

Computing the color histograms is straightforward -- just pick the range for your histogram buckets, and for each range, tally the number of pixels with a color in that range. For example, consider the "green" histogram, and suppose we choose 4 buckets for our histogram: 0-63, 64-127, 128-191, and 192-255. Then for each pixel, we look at the green value, and add a tally to the appropriate bucket. When we're done tallying, we divide each bucket total by the number of pixels in the entire image to get a normalized histogram for the green channel.

For the texture direction histogram, we started by performing edge detection on the image. Each edge point has a normal vector pointing in the direction perpendicular to the edge. We quantized the normal vector's angle into one of 6 buckets between 0 and PI (since edges have 180-degree symmetry, we converted angles between -PI and 0 to be between 0 and PI). After tallying up the number of edge points in each direction, we have an un-normalized histogram representing texture direction, which we normalized by dividing each bucket by the total number of edge points in the image.

为了计算纹理尺度直方图，对于每个边缘点，我们测量到具有相同方向的下一个最近边缘点的距离。例如，如果边缘点A的方向是45度，算法就会沿着这个方向走，直到找到另一个方向为45度的边缘点(或在合理偏差范围内)。在计算每个边缘点的距离后，我们将这些值转储到直方图中，并通过除以边缘点的总数来归一化。

现在每张图像有5个直方图。要比较两张图像，需要取每个直方图桶之间差值的绝对值，然后将这些值相加。例如，为了比较图像A和B，我们将计算

|A.green_histogram.bucket_1 - B.green_histogram.bucket_1| 

对于绿色直方图中的每个桶，并对其他直方图重复，然后将所有结果相加。结果越小，匹配越好。对数据库中的所有图像重复此操作，结果最小的匹配者获胜。您可能希望有一个阈值，超过这个阈值，算法就会得出没有找到匹配的结论。

第三个选择-关键点+决策树

A third approach that is probably much faster than the other two is using semantic texton forests (PDF). This involves extracting simple keypoints and using a collection decision trees to classify the image. This is faster than simple SIFT keypoint matching, because it avoids the costly matching process, and keypoints are much simpler than SIFT, so keypoint extraction is much faster. However, it preserves the SIFT method's invariance to rotation, scale, and lighting, an important feature that the histogram method lacked.

更新:

我的错误——语义德克顿森林论文并不是专门关于图像匹配的，而是关于区域标记的。关于匹配的原始论文是:使用随机树的关键点识别。此外，下面的论文继续发展的想法，并代表了艺术的状态(c. 2010):

快速关键点识别使用随机蕨类-更快，更可扩展比Lepetit 06 概要:二进制健壮的独立基本特征-不太健壮但非常快-我认为这里的目标是在智能手机和其他手持设备上进行实时匹配

我认为值得在此基础上添加我构建的phash解决方案，我们已经使用了一段时间:Image:: phash。它是一个Perl模块，但主要部分是用c语言编写的。它比phash.org快几倍，并且为基于dct的phash提供了一些额外的特性。

我们已经在MySQL数据库上建立了数以千万计的图像索引，所以我想要一些快速的东西，也想要一种使用MySQL索引的方法(这与汉明距离不工作)，这导致我使用“减少”哈希进行直接匹配，模块文档讨论了这一点。

使用起来很简单:

use Image::PHash;

my $iph1 = Image::PHash->new('file1.jpg');
my $p1   = $iph1->pHash();

my $iph2 = Image::PHash->new('file2.jpg');
my $p2   = $iph2->pHash();

my $diff = Image::PHash::diff($p1, $p2);

正如卡特曼所指出的，您可以使用任何类型的哈希值来查找精确的重复项。

寻找近距离图像的一个起点可能在这里。这是CG公司用来检查修改后的图像是否仍然显示本质上相同的场景的工具。

选择100个随机点可能意味着相似(有时甚至不相似)的图像将被标记为相同，我认为这不是您想要的。如果图像格式不同(png、jpeg等)、大小不同或元数据不同，MD5哈希就无法工作。将所有图像缩小到一个更小的尺寸是一个不错的选择，只要你使用的是一个好的图像库/快速的语言，做一个像素对像素的比较不应该花费太长时间，而且尺寸足够小。

你可以试着让它们变得很小，然后如果它们是一样的，就在更大的尺寸上进行另一次比较——这可能是速度和准确性的良好结合……