Java OpenCV层次结构始终为空
我有一张有重叠轮廓的图像,我一直试图在找到轮廓时使用层次过滤掉轮廓。我想做的是过滤掉双亲不等于-1的轮廓。然而,当我试图获取包含层次结构的信息时,父索引几乎每次都等于null。我是否没有查看正确的信息来获取当前家长的状态?这是我的密码Java OpenCV层次结构始终为空,java,opencv,hierarchy,Java,Opencv,Hierarchy,我有一张有重叠轮廓的图像,我一直试图在找到轮廓时使用层次过滤掉轮廓。我想做的是过滤掉双亲不等于-1的轮廓。然而,当我试图获取包含层次结构的信息时,父索引几乎每次都等于null。我是否没有查看正确的信息来获取当前家长的状态?这是我的密码 List<MatOfPoint> contours = new ArrayList<>(); List<MatOfPoint> squareContours = new ArrayList<>();
List<MatOfPoint> contours = new ArrayList<>();
List<MatOfPoint> squareContours = new ArrayList<>();
Mat hierarchy = new Mat();
//find all contours
Imgproc.findContours(dilated, contours, hierarchy, Imgproc.RETR_TREE, Imgproc.CHAIN_APPROX_SIMPLE);
//Remove contours that aren't close to a square shape.
for(int i = 0; i < contours.size(); i++){
if(hierarchy != null){
double area = Imgproc.contourArea(contours.get(i));
MatOfPoint2f contour2f = new MatOfPoint2f(contours.get(i).toArray());
double perimeter = Imgproc.arcLength(contour2f, true);
//Found squareness equation on wiki...
//https://en.wikipedia.org/wiki/Shape_factor_(image_analysis_and_microscopy)
double squareness = 4 * Math.PI * area / Math.pow(perimeter, 2);
if(squareness >= 0.7 && squareness <= 0.9 && area >= 2000){
squareContours.add(contours.get(i));
}
}
}
//remove contour if it has a parent
List<MatOfPoint> finalContours = new ArrayList<>();
for(int i = 0; i < squareContours.size();i++){
if(hierarchy.get(i, 3)[3] == -1){ //this should be checking parent index I think.
finalContours.add(squareContours.get(i));
}
}
使用
Mat
表示findContours
返回的层次结构时,会得到一个包含以下内容的数组:
- 一排
- 每个检测到的轮廓有一列
- 4个通道(下一个、上一个、子和父轮廓的id)
- 数据类型为32位有符号整数
hierarchy.get(i, 3)[3]
您使用的方法具有以下签名:
public double[] get(int row, int col)
请注意,第一个参数是行号。将等高线编号作为行传递,但只有一行
接下来,第二个参数是列。您总是会得到第3列--第3个等高线的层次信息
你真正应该做的是
hierarchy.get(0, i)[3]
最后一个问题是不必要地将索引转换为浮点数。这是一种浪费,而且适得其反,因为要发挥很大的作用,您必须将它们转换回int。只要用一个铅笔
现在,我的Java已经生锈了,但我认为您可以这样做:
int[] current_hierarchy = new int[4];
for(int i = 0; i < squareContours.size();i++) {
hierarchy.get(0, i, current_hierarchy);
if (current_hierarchy[3] == -1) {
// ... and so on
List<MatOfPoint> contours = new ArrayList<>();
Mat hierarchy = new Mat();
//find all contours
Imgproc.findContours(dilated, contours, hierarchy, Imgproc.RETR_TREE, Imgproc.CHAIN_APPROX_SIMPLE);
// Remove contours that aren't close to a square shape
// and remove contour if it has a parent
List<MatOfPoint> finalContours = new ArrayList<>();
int[] current_hierarchy = new int[4];
for(int i = 0; i < contours.size(); i++){
double area = Imgproc.contourArea(contours.get(i));
MatOfPoint2f contour2f = new MatOfPoint2f(contours.get(i).toArray());
double perimeter = Imgproc.arcLength(contour2f, true);
//Found squareness equation on wiki...
//https://en.wikipedia.org/wiki/Shape_factor_(image_analysis_and_microscopy)
double squareness = 4 * Math.PI * area / Math.pow(perimeter, 2);
if(squareness >= 0.7 && squareness <= 0.9 && area >= 2000){
hierarchy.get(0, i, current_hierarchy);
if (current_hierarchy[3] == -1) {
finalContours.add(contours.get(i));
}
}
}
List<double[]> listHierarchy = new ArrayList<>();
for (int i = 0; i< list.size(); i++){
listHierarchy.add(hierarchie.get(0, i));
}
使用
Mat
表示findContours
返回的层次结构时,会得到一个包含以下内容的数组:
- 一排
- 每个检测到的轮廓有一列
- 4个通道(下一个、上一个、子和父轮廓的id)
- 数据类型为32位有符号整数
hierarchy.get(i, 3)[3]
您使用的方法具有以下签名:
public double[] get(int row, int col)
请注意,第一个参数是行号。将等高线编号作为行传递,但只有一行
接下来,第二个参数是列。您总是会得到第3列--第3个等高线的层次信息
你真正应该做的是
hierarchy.get(0, i)[3]
最后一个问题是不必要地将索引转换为浮点数。这是一种浪费,而且适得其反,因为要发挥很大的作用,您必须将它们转换回int。只要用一个铅笔
现在,我的Java已经生锈了,但我认为您可以这样做:
int[] current_hierarchy = new int[4];
for(int i = 0; i < squareContours.size();i++) {
hierarchy.get(0, i, current_hierarchy);
if (current_hierarchy[3] == -1) {
// ... and so on
List<MatOfPoint> contours = new ArrayList<>();
Mat hierarchy = new Mat();
//find all contours
Imgproc.findContours(dilated, contours, hierarchy, Imgproc.RETR_TREE, Imgproc.CHAIN_APPROX_SIMPLE);
// Remove contours that aren't close to a square shape
// and remove contour if it has a parent
List<MatOfPoint> finalContours = new ArrayList<>();
int[] current_hierarchy = new int[4];
for(int i = 0; i < contours.size(); i++){
double area = Imgproc.contourArea(contours.get(i));
MatOfPoint2f contour2f = new MatOfPoint2f(contours.get(i).toArray());
double perimeter = Imgproc.arcLength(contour2f, true);
//Found squareness equation on wiki...
//https://en.wikipedia.org/wiki/Shape_factor_(image_analysis_and_microscopy)
double squareness = 4 * Math.PI * area / Math.pow(perimeter, 2);
if(squareness >= 0.7 && squareness <= 0.9 && area >= 2000){
hierarchy.get(0, i, current_hierarchy);
if (current_hierarchy[3] == -1) {
finalContours.add(contours.get(i));
}
}
}
List<double[]> listHierarchy = new ArrayList<>();
for (int i = 0; i< list.size(); i++){
listHierarchy.add(hierarchie.get(0, i));
}
从层级列表中删除项目时,还需要检查已删除项目的关系。我首先将完整的层次结构矩阵传输到如下列表:
int[] current_hierarchy = new int[4];
for(int i = 0; i < squareContours.size();i++) {
hierarchy.get(0, i, current_hierarchy);
if (current_hierarchy[3] == -1) {
// ... and so on
List<MatOfPoint> contours = new ArrayList<>();
Mat hierarchy = new Mat();
//find all contours
Imgproc.findContours(dilated, contours, hierarchy, Imgproc.RETR_TREE, Imgproc.CHAIN_APPROX_SIMPLE);
// Remove contours that aren't close to a square shape
// and remove contour if it has a parent
List<MatOfPoint> finalContours = new ArrayList<>();
int[] current_hierarchy = new int[4];
for(int i = 0; i < contours.size(); i++){
double area = Imgproc.contourArea(contours.get(i));
MatOfPoint2f contour2f = new MatOfPoint2f(contours.get(i).toArray());
double perimeter = Imgproc.arcLength(contour2f, true);
//Found squareness equation on wiki...
//https://en.wikipedia.org/wiki/Shape_factor_(image_analysis_and_microscopy)
double squareness = 4 * Math.PI * area / Math.pow(perimeter, 2);
if(squareness >= 0.7 && squareness <= 0.9 && area >= 2000){
hierarchy.get(0, i, current_hierarchy);
if (current_hierarchy[3] == -1) {
finalContours.add(contours.get(i));
}
}
}
List<double[]> listHierarchy = new ArrayList<>();
for (int i = 0; i< list.size(); i++){
listHierarchy.add(hierarchie.get(0, i));
}
List listHierarchy=new ArrayList();
对于(int i=0;i
然后,当我需要从此列表中删除项目时,我调用此函数:
List<double[]> deleteHierarchyItem(int position, List<double[]> hierarchy){
double[] itemHierarchy = hierarchy.get(position);
double[] workItem;
//doesnt have children?
if (itemHierarchy[2]!=-1){
int nextChild =(int) (itemHierarchy[2]);
do {
//nacte dite
workItem = hierarchy.get(nextChild);
//zmeni rodice v diteti na rodice puvodniho rodice
workItem[3] = itemHierarchy[3];
//nastavi nova data v listu
hierarchy.set(nextChild, workItem);
//zmeni ukazatel na nove dite
nextChild = (int)(workItem[2]);
} while (nextChild != -1);
}
//check for siblings
boolean hasNextSibling = itemHierarchy[0] != -1;
boolean hasPreviousSibling = itemHierarchy[1] != -1;
double idPreviousSibling = itemHierarchy[1];
double idNextSibling = itemHierarchy[0];
//has both siblings
if (hasPreviousSibling && hasNextSibling){
//change previous sibling
workItem = hierarchy.get((int)(idPreviousSibling));
workItem[0] = idNextSibling;
hierarchy.set((int)(idPreviousSibling), workItem);
//change next sibling
workItem = hierarchy.get((int)(idNextSibling));
workItem[1] = idPreviousSibling;
hierarchy.set((int)(idNextSibling), workItem);
}
//has only previous sibling
if (hasPreviousSibling && !hasNextSibling){
workItem = hierarchy.get((int)(idPreviousSibling));
workItem[0] = -1;
hierarchy.set((int)(idPreviousSibling), workItem);
}
//has only next sibling
if (!hasPreviousSibling && hasNextSibling){
workItem = hierarchy.get((int)(idNextSibling));
workItem[1] = -1;
hierarchy.set((int)(idNextSibling), workItem);
//change of child parametres in parent
if(itemHierarchy[3]>0)
{
workItem = hierarchy.get((int)(itemHierarchy[3]));
workItem[2]=idNextSibling;
hierarchy.set((int)(itemHierarchy[3]), workItem);
}
}
//check for parent
if (itemHierarchy[3]!=-1){
workItem = hierarchy.get((int)(itemHierarchy[3]));
if (workItem[2]==position){
workItem[2] = -1;
hierarchy.set((int)(itemHierarchy[3]), workItem);
}
}
//iterate and decrement values
for (int i = position; i< hierarchy.size();i++){
workItem = hierarchy.get(i);
if (workItem[0]>position){
workItem[0] = workItem[0] - 1;
}
if (workItem[1]>position){
workItem[1] = workItem[1] - 1;
}
if (workItem[2]>position){
workItem[2] = workItem[2] - 1;
}
if (workItem[3]>position){
workItem[3] = workItem[3] - 1;
}
hierarchy.set(i, workItem);
}
return hierarchy;
}
List deleteHierarchyItem(int位置,列表层次结构){
double[]itemHierarchy=hierarchy.get(位置);
双[]工作项;
//没有孩子吗?
如果(itemHierarchy[2]!=-1){
int nextChild=(int)(itemHierarchy[2]);
做{
//钠碲铋矿
workItem=hierarchy.get(nextChild);
//zmeni rodice v Dieti na rodice puvodniho rodice
工作项[3]=项目层次结构[3];
//纳斯塔维新星数据v列表
hierarchy.set(nextChild,workItem);
//zmeni ukazatel na nove酒店
nextChild=(int)(工作项[2]);
}while(nextChild!=-1);
}
//检查兄弟姐妹
布尔hasNextSibling=itemHierarchy[0]!=-1;
布尔hasPreviousSibling=itemHierarchy[1]!=-1;
双idPreviousSibling=itemHierarchy[1];
double idNextSibling=itemHierarchy[0];
//他有两个兄弟姐妹
if(hasPreviousSibling和&hasNextSibling){
//改变以前的兄弟姐妹
workItem=hierarchy.get((int)(idPreviousSibling));
工作项[0]=idNextSibling;
集合((int)(idPreviousSibling),workItem);
//换下一个兄弟姐妹
workItem=hierarchy.get((int)(idNextSibling));
工作项[1]=idPreviousSibling;
集合((int)(idNextSibling),workItem);
}
//他只有以前的兄弟姐妹
if(hasPreviousSibling&!hasNextSibling){
workItem=hierarchy.get((int)(idPreviousSibling));
工作项[0]=-1;
集合((int)(idPreviousSibling),workItem);
}
//他只有下一个兄弟姐妹
如果(!hasPreviousSibling&&hasNextSibling){
workItem=hierarchy.get((int)(idNextSibling));
工作项[1]=-1;
集合((int)(idNextSibling),workItem);
//更改父级中的子参数
如果(itemHierarchy[3]>0)
{
workItem=hierarchy.get((int)(itemHierarchy[3]);
工作项[2]=idNextSibling;
集合((int)(itemHierarchy[3]),workItem);
}
}
//检查家长
如果(itemHierarchy[3]!=-1){
workItem=hierarchy.get((int)(itemHierarchy[3]);
if(工作项[2]==位置){
工作项[2]=-1;
集合((int)(itemHierarchy[3]),workItem);
}
}
//迭代和递减值
for(int i=position;i位置){
工作项[0]=工作项[0]-1;
}
if(工作项[1]>位置){
工作项[1]=工作项[1]-1;
}
if(工作项[2]>位置){
工作项[2]=工作项[2]-1;
}
if(工作项[3]>位置){
工作项[3]=工作项[3]-1;
}
集合(i,工作项);
}
返回层次;
}
从层级列表中删除项目时,还需要检查已删除项目的关系。我首先将完整的层次结构表传输到如下列表中