我的任务是使用OpenCV和c ++创建一个应用程序，它将接收植物叶子的图像输入。该应用将检测可能的疾病症状，例如来自叶子的黑色/灰色/褐色斑点，或枯萎病，病变等。疾病的每种特征（例如斑点的颜色）代表不同的疾病。在检测到可能的症状后，应用程序将其与应用程序数据库中的模板图像集合进行匹配，并输出可能的最佳匹配。
我必须使用哪些方法？我已经研究了直方图匹配和关键点和描述符匹配，但我不确定哪一个最适合。
我找到了使用SURF和FLANN的示例代码，但我不知道这是否足够：
#include <stdio.h>
#include <iostream>
#include "opencv2/core/core.hpp"
#include "opencv2/features2d/features2d.hpp"
#include "opencv2/highgui/highgui.hpp"
#include "opencv2/nonfree/features2d.hpp"
using namespace cv;
void readme();
/**
* @function main
* @brief Main function
*/
int main( int argc, char** argv )
{
if( argc != 3 )
{ readme(); return -1; }
Mat img_1 = imread( argv[1], CV_LOAD_IMAGE_GRAYSCALE );
Mat img_2 = imread( argv[2], CV_LOAD_IMAGE_GRAYSCALE );
if( !img_1.data || !img_2.data )
{ std::cout<< " --(!) Error reading images " << std::endl; return -1; }
//-- Step 1: Detect the keypoints using SURF Detector
int minHessian = 400;
SurfFeatureDetector detector( minHessian );
std::vector<KeyPoint> keypoints_1, keypoints_2;
detector.detect( img_1, keypoints_1 );
detector.detect( img_2, keypoints_2 );
//-- Step 2: Calculate descriptors (feature vectors)
SurfDescriptorExtractor extractor;
Mat descriptors_1, descriptors_2;
extractor.compute( img_1, keypoints_1, descriptors_1 );
extractor.compute( img_2, keypoints_2, descriptors_2 );
//-- Step 3: Matching descriptor vectors using FLANN matcher
FlannBasedMatcher matcher;
std::vector< DMatch > matches;
matcher.match( descriptors_1, descriptors_2, matches );
double max_dist = 0; double min_dist = 100;
//-- Quick calculation of max and min distances between keypoints
for( int i = 0; i < descriptors_1.rows; i++ )
{ double dist = matches[i].distance;
if( dist < min_dist ) min_dist = dist;
if( dist > max_dist ) max_dist = dist;
}
printf("-- Max dist : %f \n", max_dist );
printf("-- Min dist : %f \n", min_dist );
//-- Draw only "good" matches (i.e. whose distance is less than 2*min_dist,
//-- or a small arbitary value ( 0.02 ) in the event that min_dist is very
//-- small)
//-- PS.- radiusMatch can also be used here.
std::vector< DMatch > good_matches;
for( int i = 0; i < descriptors_1.rows; i++ )
{ if( matches[i].distance <= max(2*min_dist, 0.02) )
{ good_matches.push_back( matches[i]); }
}
//-- Draw only "good" matches
Mat img_matches;
drawMatches( img_1, keypoints_1, img_2, keypoints_2,
good_matches, img_matches, Scalar::all(-1), Scalar::all(-1),
vector<char>(), DrawMatchesFlags::NOT_DRAW_SINGLE_POINTS );
//-- Show detected matches
imshow( "Good Matches", img_matches );
for( int i = 0; i < (int)good_matches.size(); i++ )
{ printf( "-- Good Match [%d] Keypoint 1: %d -- Keypoint 2: %d \n", i, good_matches[i].queryIdx, good_matches[i].trainIdx ); }
waitKey(0);
return 0;
}
/**
* @function readme
*/
void readme()
{ std::cout << " Usage: ./SURF_FlannMatcher <img1> <img2>" << std::endl; }
以下是我的问题：
我必须使用哪种方法？直方图匹配，关键点/描述符匹配或？
如果我使用Keypoint / Descriptor匹配，什么算法是SURF和FLANN的最佳替代方法，因为我将在Android平台上实现它？我还需要进行阈值处理或分割吗？它是否会删除重要的细节，如颜色，形状等？请伙计们，建议采取一些措施。
答案 0 :(得分：1)
我认为这种方式应该会给你带来好结果：
培训过程。
预测过程：
我成功地将前馈神经网络用作分类器，以解决类似的问题。
您可能会发现本书很有用：ISBN 978-0-85729-747-1＆＃34;使用本地二进制模式的计算机视觉＆＃34;
尝试这个（计算LBP描述符，还有计算直方图的函数）：
#include <iostream>
#include <opencv2/opencv.hpp>
#include <opencv2/imgproc/imgproc.hpp>
#include <opencv2/highgui/highgui.hpp>
#include <opencv2/features2d/features2d.hpp>
#include "opencv2/nonfree/nonfree.hpp"
#include <limits>
using namespace cv;
class myLBP
{
public:
uchar lut[256];
uchar null;
int radius;
int maxTransitions;
bool rotationInvariant;
myLBP(int _radius=1,int _maxTransitions=8,bool _rotationInvariant=false)
{
radius=_radius;
maxTransitions=_maxTransitions;
rotationInvariant=_rotationInvariant;
bool set[256];
uchar uid = 0;
for (int i=0; i<256; i++)
{
if (numTransitions(i) <= maxTransitions)
{
int id;
if (rotationInvariant)
{
int rie = rotationInvariantEquivalent(i);
if (i == rie)
{
id = uid++;
}
else
{
id = lut[rie];
}
}
else
{
id = uid++;
}
lut[i] = id;
set[i] = true;
}
else
{
set[i] = false;
}
}
null = uid;
for (int i=0; i<256; i++)
if (!set[i])
{
lut[i] = null; // Set to null id
}
}
/* Returns the number of 0->1 or 1->0 transitions in i */
static int numTransitions(int i)
{
int transitions = 0;
int curParity = i%2;
for (int j=1; j<=8; j++)
{
int parity = (i>>(j%8)) % 2;
if (parity != curParity)
{
transitions++;
}
curParity = parity;
}
return transitions;
}
static int rotationInvariantEquivalent(int i)
{
int min = std::numeric_limits<int>::max();
for (int j=0; j<8; j++)
{
bool parity = i % 2;
i = i >> 1;
if (parity)
{
i+=128;
}
min = std::min(min, i);
}
return min;
}
void process(const Mat &src, Mat &dst) const
{
Mat m;
src.convertTo(m, CV_32F);
assert(m.isContinuous() && (m.channels() == 1));
Mat n(m.rows, m.cols, CV_8UC1);
n = null; // Initialize to NULL LBP pattern
const float *p = (const float*)m.ptr();
for (int r=radius; r<m.rows-radius; r++)
{
for (int c=radius; c<m.cols-radius; c++)
{
const float cval = (p[(r+0*radius)*m.cols+c+0*radius]);
n.at<uchar>(r, c) = lut[(p[(r-1*radius)*m.cols+c-1*radius] >= cval ? 128 : 0) |
(p[(r-1*radius)*m.cols+c+0*radius] >= cval ? 64 : 0) |
(p[(r-1*radius)*m.cols+c+1*radius] >= cval ? 32 : 0) |
(p[(r+0*radius)*m.cols+c+1*radius] >= cval ? 16 : 0) |
(p[(r+1*radius)*m.cols+c+1*radius] >= cval ? 8 : 0) |
(p[(r+1*radius)*m.cols+c+0*radius] >= cval ? 4 : 0) |
(p[(r+1*radius)*m.cols+c-1*radius] >= cval ? 2 : 0) |
(p[(r+0*radius)*m.cols+c-1*radius] >= cval ? 1 : 0)];
}
}
dst=n.clone();
}
/* Returns the number of 1 bits in i */
static int bitCount(int i)
{
int count = 0;
for (int j=0; j<8; j++)
{
count += (i>>j)%2;
}
return count;
}
void draw(const Mat &src, Mat &dst) const
{
static Mat hueLUT, saturationLUT, valueLUT;
if (!hueLUT.data)
{
const int NUM_COLORS = 10;
hueLUT.create(1, 256, CV_8UC1);
hueLUT.setTo(0);
uchar uid = 0;
for (int i=0; i<256; i++)
{
const int transitions = numTransitions(i);
int u2;
if (transitions <= 2)
{
u2 = uid++;
}
else
{
u2 = 58;
}
// Assign hue based on bit count
int color = bitCount(i);
if (transitions > 2)
{
color = NUM_COLORS-1;
}
hueLUT.at<uchar>(0, u2) = 255.0*(float)color/(float)NUM_COLORS;
}
saturationLUT.create(1, 256, CV_8UC1);
saturationLUT.setTo(255);
valueLUT.create(1, 256, CV_8UC1);
valueLUT.setTo(255.0*(3.0/4.0));
}
if (src.type() != CV_8UC1)
{
std::cout << "Expected 8UC1 source type.";
}
Mat hue, saturation, value;
LUT(src, hueLUT, hue);
LUT(src, saturationLUT, saturation);
LUT(src, valueLUT, value);
std::vector<Mat> mv;
mv.push_back(hue);
mv.push_back(saturation);
mv.push_back(value);
Mat coloredU2;
merge(mv, coloredU2);
cvtColor(coloredU2, dst, cv::COLOR_HSV2BGR);
}
};
void Hist(const Mat &src, Mat &dst,float max=256, float min=0,int dims=-1)
{
std::vector<Mat> mv;
split(src, mv);
Mat m(mv.size(), dims, CV_32FC1);
for (size_t i=0; i<mv.size(); i++)
{
int channels[] = {0};
int histSize[] = {dims};
float range[] = {min, max};
const float* ranges[] = {range};
Mat hist, chan = mv[i];
// calcHist requires F or U, might as well convert just in case
if (mv[i].depth() != CV_8U && mv[i].depth() != CV_32F)
{
mv[i].convertTo(chan, CV_32F);
}
calcHist(&chan, 1, channels, Mat(), hist, 1, histSize, ranges);
memcpy(m.ptr(i), hist.ptr(), dims * sizeof(float));
}
dst=m.clone();
}
int main(int argc, char* argv[])
{
cv::initModule_nonfree();
cv::namedWindow("result");
cv::Mat bgr_img = cv::imread("D:\\ImagesForTest\\lena.jpg");
if (bgr_img.empty())
{
exit(EXIT_FAILURE);
}
cv::Mat gray_img;
cv::cvtColor(bgr_img, gray_img, cv::COLOR_BGR2GRAY);
cv::normalize(gray_img, gray_img, 0, 255, cv::NORM_MINMAX);
myLBP lbp(1,2);
Mat lbp_img;
lbp.process(gray_img,lbp_img);
lbp.draw(lbp_img,bgr_img);
//for(int i=0;i<lbp_img.rows;++i)
imshow("result",bgr_img);
cv::waitKey();
return 0;
}