在c++中,我正在这样做:一切都很好,现在我需要切换到java for android。
vector<vector<Point> > contours;
findContours(image, contours,
CV_RETR_EXTERNAL,CV_CHAIN_APPROX_SIMPLE);
printf( "\n%d contours\n\n", contours.size() ) ;
/* our ASSUMPTION is that the contour with the most points is
the one we want */
int mx = 0;
int nm = -1;
for( int i = 0; i < contours.size(); i++ ){
if( contours[i].size() > mx ) {
mx = contours[i].size() ; nm = i ;
}
}
printf( "largest contour (number %d) has %d points.\n\n",
nm, mx );
Point ul, ur, lr, ll;
ul = contours[nm][0];
for( int i = 1; i < contours[nm].size(); i++ ) {
/* TODO -- handle equal case */
if( ( contours[nm][i].x + contours[nm][i].y ) <
( ul.x + ul.y ) ){ ul = contours[nm][i]; }
if( ( contours[nm][i].x + contours[nm][i].y ) >
( lr.x + lr.y ) ){ lr = contours[nm][i]; }
if( ( contours[nm][i].x - contours[nm][i].y ) >
( ur.x - ur.y ) ){ ur = contours[nm][i]; }
if( ( contours[nm][i].x - contours[nm][i].y ) <
( ll.x - ll.y ) ){ ll = contours[nm][i]; }
}
printf( "The upper left point is at ( %d, %d )\n\n",
ul.x , ul.y );
printf( "The upper right point is at ( %d, %d )\n\n",
ur.x , ur.y );
printf( "The lower right point is at ( %d, %d )\n\n",
lr.x , lr.y );
printf( "The lower left point is at ( %d, %d )\n\n",
ll.x , ll.y );
/* got the corners */
java使用了一个“matofpoint”列表,因此我无法获得与每个轮廓相关联的点的数量。
我觉得不应该这么难。
List<MatOfPoint > contours = null;
Mat hierarchy = new Mat();
findContours( grayimage, contours, hierarchy, Imgproc.RETR_LIST,
Imgproc.CHAIN_APPROX_SIMPLE);
for ( int i = 0 ; i < contours.size() ; i++ ) {
int sss = contours.get( i ).size() ;
}
但是contours.get(i).size()返回的size对象不是我想要的。
非常感谢您的帮助。
谢谢,
吉姆
1条答案
按热度按时间uqdfh47h1#
int sss=等高线.get(i).tolist().size();