21

我找到了一些关于通过 OpenCV 拼接全景图像的基本工作示例。我还在API 文档中找到了一些有用的文档,但我无法通过提供其他信息来了解如何加快处理速度。

在我的例子中,我在单个帧的 20x20 网格中生成一组图像,总共将 400 个图像拼接成一个大的图像。这在现代 PC 上需要花费大量时间,因此在开发板上可能需要数小时。

有没有办法告诉 OpenCV 实例有关图像的信息,例如我提前知道所有图像的相对位置,因为它们会出现在网格上?到目前为止,我看到的唯一 API 调用是通过vImg.push_back().


参考

  1. 缝合。图像拼接 - OpenCV API 文档,访问时间:2014-02-26,<http://docs.opencv.org/modules/stitching/doc/stitching.html>
  2. OpenCV 拼接示例(Stitcher 类,全景图),访问时间:2014-02-26,<http://feelmare.blogspot.ca/2013/11/opencv-stitching-example-stitcher-class.html>
  3. Panorama – OpenCV 中的图像拼接,访问时间:2014-02-26,<http://ramsrigoutham.com/2012/11/22/panorama-image-stitching-in-opencv/>
4

5 回答 5

9

我对拼接管道做了一些工作,虽然我不认为自己是该领域的专家,但我确实获得了更好的性能(以及更好的结果)分别调整管道的每个步骤。正如你在图片中看到的,Stitching 类只不过是这个管道的一个包装器:拼接管道概述

您可以调整的一些有趣的部分是调整大小的步骤(有一点是更高的分辨率只是意味着更多的计算时间和更不准确的特征),匹配过程和(尽管这只是一个猜测)提供一个好的相机参数而不是执行估计。这涉及在进行拼接之前获取相机参数,但这并不难。在这里你有一些参考:OpenCV Camera Calibration and 3D Reconstruction

再说一遍:我不是专家,这只是基于我作为实习生对图书馆进行一些实验的经验!

于 2015-04-06T10:02:35.430 回答
7

据我所知,除了给它一个图像列表之外,没有办法向 OpenCV 引擎提供额外的数据。不过,它本身就做得很好。我会查看一些示例代码,并测试每个拼接操作需要多长时间。从我使用 4x6、4x8、...、4x20 全景重建的实验来看,所需的 CPU 时间似乎随着重叠图像的数量而增加。我想您的案例至少需要一分钟才能在现代机器上进行计算。

来源: https ://code.ros.org/trac/opencv/browser/trunk/opencv/samples/cpp/stitching.cpp?rev=6682

1   /*M///////////////////////////////////////////////////////////////////////////////////////
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10  //                          License Agreement
11  //                For Open Source Computer Vision Library
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26  //
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28  //     derived from this software without specific prior written permission.
29  //
30  // This software is provided by the copyright holders and contributors "as is" and
31  // any express or implied warranties, including, but not limited to, the implied
32  // warranties of merchantability and fitness for a particular purpose are disclaimed.
33  // In no event shall the Intel Corporation or contributors be liable for any direct,
34  // indirect, incidental, special, exemplary, or consequential damages
35  // (including, but not limited to, procurement of substitute goods or services;
36  // loss of use, data, or profits; or business interruption) however caused
37  // and on any theory of liability, whether in contract, strict liability,
38  // or tort (including negligence or otherwise) arising in any way out of
39  // the use of this software, even if advised of the possibility of such damage.
40  //
41  //M*/
42  
43  // We follow to these papers:
44  // 1) Construction of panoramic mosaics with global and local alignment.
45  //    Heung-Yeung Shum and Richard Szeliski. 2000.
46  // 2) Eliminating Ghosting and Exposure Artifacts in Image Mosaics.
47  //    Matthew Uyttendaele, Ashley Eden and Richard Szeliski. 2001.
48  // 3) Automatic Panoramic Image Stitching using Invariant Features.
49  //    Matthew Brown and David G. Lowe. 2007.
50  
51  #include <iostream>
52  #include <fstream>
53  #include "opencv2/highgui/highgui.hpp"
54  #include "opencv2/stitching/stitcher.hpp"
55  
56  using namespace std;
57  using namespace cv;
58  
59  void printUsage()
60  {
61      cout <<
62          "Rotation model images stitcher.\n\n"
63          "stitching img1 img2 [...imgN]\n\n"
64          "Flags:\n"
65          "  --try_use_gpu (yes|no)\n"
66          "      Try to use GPU. The default value is 'no'. All default values\n"
67          "      are for CPU mode.\n"
68          "  --output <result_img>\n"
69          "      The default is 'result.jpg'.\n";
70  }
71  
72  bool try_use_gpu = false;
73  vector<Mat> imgs;
74  string result_name = "result.jpg";
75  
76  int parseCmdArgs(int argc, char** argv)
77  {
78      if (argc == 1)
79      {
80          printUsage();
81          return -1;
82      }
83      for (int i = 1; i < argc; ++i)
84      {
85          if (string(argv[i]) == "--help" || string(argv[i]) == "/?")
86          {
87              printUsage();
88              return -1;
89          }
90          else if (string(argv[i]) == "--try_gpu")
91          {
92              if (string(argv[i + 1]) == "no")
93                  try_use_gpu = false;
94              else if (string(argv[i + 1]) == "yes")
95                  try_use_gpu = true;
96              else
97              {
98                  cout << "Bad --try_use_gpu flag value\n";
99                  return -1;
100             }
101             i++;
102         }
103         else if (string(argv[i]) == "--output")
104         {
105             result_name = argv[i + 1];
106             i++;
107         }
108         else
109         {
110             Mat img = imread(argv[i]);
111             if (img.empty())
112             {
113                 cout << "Can't read image '" << argv[i] << "'\n";
114                 return -1;
115             }
116             imgs.push_back(img);
117         }
118     }
119     return 0;
120 }
121 
122 
123 int main(int argc, char* argv[])
124 {
125     int retval = parseCmdArgs(argc, argv);
126     if (retval) return -1;
127 
128     Mat pano;
129     Stitcher stitcher = Stitcher::createDefault(try_use_gpu);
130     Stitcher::Status status = stitcher.stitch(imgs, pano);
131 
132     if (status != Stitcher::OK)
133     {
134         cout << "Can't stitch images, error code = " << status << endl;
135         return -1;
136     }
137 
138     imwrite(result_name, pano);
139     return 0;
140 }
141 
142 
于 2014-03-11T15:32:24.533 回答
4

也许这会有所帮助? https://software.intel.com/en-us/articles/fast-panorama-stitching

特别是关于成对匹配的部分

罗南

于 2014-07-13T14:10:22.650 回答
3

考虑在 Opencv Stitcher 中启用 GPU:

bool try_use_gpu = true;
Stitcher myStitcher = Stitcher::createDefault(try_use_gpu); 
Stitcher::Status status = myStitcher.stitch(Imgs, pano);
于 2014-04-03T19:21:52.537 回答
2

如果您知道图像的相对位置,您似乎可以将问题分解为子问题,并可能通过了解问题的子结构来减少计算量。基本上将图像集分成 4 个相邻图像的组,处理帧,然后使用相同的想法继续处理生成的图像,直到您到达全景图。话虽如此,我最近才开始玩弄这个 opencv 工具集。我知道这是一个非常简单的想法,但它可能对某人有用。

于 2015-11-24T08:39:16.573 回答