Doxygen tutorials: basic structure
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Maksim Shabunin
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Histogram Comparison {#tutorial_histogram_comparison}
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====================
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Goal
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----
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In this tutorial you will learn how to:
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- Use the function @ref cv::compareHist to get a numerical parameter that express how well two
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histograms match with each other.
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- Use different metrics to compare histograms
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Theory
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------
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- To compare two histograms ( \f$H_{1}\f$ and \f$H_{2}\f$ ), first we have to choose a *metric*
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(\f$d(H_{1}, H_{2})\f$) to express how well both histograms match.
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- OpenCV implements the function @ref cv::compareHist to perform a comparison. It also offers 4
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different metrics to compute the matching:
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a. **Correlation ( CV_COMP_CORREL )**
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\f[d(H_1,H_2) = \frac{\sum_I (H_1(I) - \bar{H_1}) (H_2(I) - \bar{H_2})}{\sqrt{\sum_I(H_1(I) - \bar{H_1})^2 \sum_I(H_2(I) - \bar{H_2})^2}}\f]
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where
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\f[\bar{H_k} = \frac{1}{N} \sum _J H_k(J)\f]
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and \f$N\f$ is the total number of histogram bins.
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b. **Chi-Square ( CV_COMP_CHISQR )**
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\f[d(H_1,H_2) = \sum _I \frac{\left(H_1(I)-H_2(I)\right)^2}{H_1(I)}\f]
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c. **Intersection ( method=CV_COMP_INTERSECT )**
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\f[d(H_1,H_2) = \sum _I \min (H_1(I), H_2(I))\f]
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d. **Bhattacharyya distance ( CV_COMP_BHATTACHARYYA )**
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\f[d(H_1,H_2) = \sqrt{1 - \frac{1}{\sqrt{\bar{H_1} \bar{H_2} N^2}} \sum_I \sqrt{H_1(I) \cdot H_2(I)}}\f]
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Code
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----
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- **What does this program do?**
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- Loads a *base image* and 2 *test images* to be compared with it.
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- Generate 1 image that is the lower half of the *base image*
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- Convert the images to HSV format
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- Calculate the H-S histogram for all the images and normalize them in order to compare them.
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- Compare the histogram of the *base image* with respect to the 2 test histograms, the
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histogram of the lower half base image and with the same base image histogram.
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- Display the numerical matching parameters obtained.
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- **Downloadable code**: Click
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[here](https://github.com/Itseez/opencv/tree/master/samples/cpp/tutorial_code/Histograms_Matching/compareHist_Demo.cpp)
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- **Code at glance:**
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@includelineno cpp/tutorial_code/Histograms_Matching/compareHist_Demo.cpp
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Explanation
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-----------
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1. Declare variables such as the matrices to store the base image and the two other images to
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compare ( RGB and HSV )
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@code{.cpp}
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Mat src_base, hsv_base;
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Mat src_test1, hsv_test1;
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Mat src_test2, hsv_test2;
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Mat hsv_half_down;
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@endcode
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2. Load the base image (src_base) and the other two test images:
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@code{.cpp}
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if( argc < 4 )
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{ printf("** Error. Usage: ./compareHist_Demo <image_settings0> <image_setting1> <image_settings2>\n");
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return -1;
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}
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src_base = imread( argv[1], 1 );
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src_test1 = imread( argv[2], 1 );
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src_test2 = imread( argv[3], 1 );
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@endcode
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3. Convert them to HSV format:
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@code{.cpp}
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cvtColor( src_base, hsv_base, COLOR_BGR2HSV );
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cvtColor( src_test1, hsv_test1, COLOR_BGR2HSV );
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cvtColor( src_test2, hsv_test2, COLOR_BGR2HSV );
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@endcode
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4. Also, create an image of half the base image (in HSV format):
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@code{.cpp}
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hsv_half_down = hsv_base( Range( hsv_base.rows/2, hsv_base.rows - 1 ), Range( 0, hsv_base.cols - 1 ) );
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@endcode
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5. Initialize the arguments to calculate the histograms (bins, ranges and channels H and S ).
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@code{.cpp}
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int h_bins = 50; int s_bins = 60;
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int histSize[] = { h_bins, s_bins };
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float h_ranges[] = { 0, 180 };
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float s_ranges[] = { 0, 256 };
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const float* ranges[] = { h_ranges, s_ranges };
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int channels[] = { 0, 1 };
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@endcode
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6. Create the MatND objects to store the histograms:
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@code{.cpp}
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MatND hist_base;
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MatND hist_half_down;
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MatND hist_test1;
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MatND hist_test2;
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@endcode
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7. Calculate the Histograms for the base image, the 2 test images and the half-down base image:
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@code{.cpp}
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calcHist( &hsv_base, 1, channels, Mat(), hist_base, 2, histSize, ranges, true, false );
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normalize( hist_base, hist_base, 0, 1, NORM_MINMAX, -1, Mat() );
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calcHist( &hsv_half_down, 1, channels, Mat(), hist_half_down, 2, histSize, ranges, true, false );
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normalize( hist_half_down, hist_half_down, 0, 1, NORM_MINMAX, -1, Mat() );
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calcHist( &hsv_test1, 1, channels, Mat(), hist_test1, 2, histSize, ranges, true, false );
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normalize( hist_test1, hist_test1, 0, 1, NORM_MINMAX, -1, Mat() );
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calcHist( &hsv_test2, 1, channels, Mat(), hist_test2, 2, histSize, ranges, true, false );
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normalize( hist_test2, hist_test2, 0, 1, NORM_MINMAX, -1, Mat() );
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@endcode
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8. Apply sequentially the 4 comparison methods between the histogram of the base image (hist_base)
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and the other histograms:
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@code{.cpp}
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for( int i = 0; i < 4; i++ )
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{ int compare_method = i;
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double base_base = compareHist( hist_base, hist_base, compare_method );
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double base_half = compareHist( hist_base, hist_half_down, compare_method );
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double base_test1 = compareHist( hist_base, hist_test1, compare_method );
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double base_test2 = compareHist( hist_base, hist_test2, compare_method );
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printf( " Method [%d] Perfect, Base-Half, Base-Test(1), Base-Test(2) : %f, %f, %f, %f \n", i, base_base, base_half , base_test1, base_test2 );
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}
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@endcode
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Results
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-------
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1. We use as input the following images:
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----------- ----------- -----------
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|Base_0| |Test_1| |Test_2|
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----------- ----------- -----------
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where the first one is the base (to be compared to the others), the other 2 are the test images.
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We will also compare the first image with respect to itself and with respect of half the base
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image.
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2. We should expect a perfect match when we compare the base image histogram with itself. Also,
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compared with the histogram of half the base image, it should present a high match since both
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are from the same source. For the other two test images, we can observe that they have very
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different lighting conditions, so the matching should not be very good:
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3. Here the numeric results:
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*Method* Base - Base Base - Half Base - Test 1 Base - Test 2
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----------------- ------------- ------------- --------------- ---------------
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*Correlation* 1.000000 0.930766 0.182073 0.120447
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*Chi-square* 0.000000 4.940466 21.184536 49.273437
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*Intersection* 24.391548 14.959809 3.889029 5.775088
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*Bhattacharyya* 0.000000 0.222609 0.646576 0.801869
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For the *Correlation* and *Intersection* methods, the higher the metric, the more accurate the
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match. As we can see, the match *base-base* is the highest of all as expected. Also we can observe
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that the match *base-half* is the second best match (as we predicted). For the other two metrics,
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the less the result, the better the match. We can observe that the matches between the test 1 and
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test 2 with respect to the base are worse, which again, was expected.
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