How to send cluster in separated node ros pcl - ros

Hi i'm new in pointcloud library. I'm trying to show clustering result point on rviz or pcl viewer, and then show nothing. And i realize that my data show nothing too when i subcsribe and cout that. Hopefully can help my problem, thanks
This is my code for clustering and send node
void cloudReceive(const sensor_msgs::PointCloud2ConstPtr& inputMsg){
mutex_lock.lock();
pcl::fromROSMsg(*inputMsg, *inputCloud);
cout<<inputCloud<<endl;
pcl::search::KdTree<pcl::PointXYZRGB>::Ptr tree (new pcl::search::KdTree<pcl::PointXYZRGB>);
tree->setInputCloud(inputCloud);
std::vector<pcl::PointIndices> cluster_indices;
pcl::EuclideanClusterExtraction<pcl::PointXYZRGB> ec;
ec.setClusterTolerance(0.03);//2cm
ec.setMinClusterSize(200);//min points
ec.setMaxClusterSize(1000);//max points
ec.setSearchMethod(tree);
ec.setInputCloud(inputCloud);
ec.extract(cluster_indices);
if(cluster_indices.size() > 0){
std::vector<pcl::PointIndices>::const_iterator it;
int i = 0;
for (it = cluster_indices.begin(); it != cluster_indices.end(); ++it){
if(i >= 10)
break;
cloud_cluster[i]->points.clear();
std::vector<int>::const_iterator idx_it;
for (idx_it = it->indices.begin(); idx_it != it->indices.end(); idx_it++)
cloud_cluster[i]->points.push_back(inputCloud->points[*idx_it]);
cloud_cluster[i]->width = cloud_cluster[i]->points.size();
// cloud_cluster[i]->height = 1;
// cloud_cluster[i]->is_dense = true;
cout<<"PointCloud representing the Cluster: " << cloud_cluster[i]->points.size() << " data points"<<endl;
std::stringstream ss;
ss<<"cobaa_pipecom2_cluster_"<< i << ".pcd";
writer.write<pcl::PointXYZRGB> (ss.str(), *cloud_cluster[i], false);
pcl::toROSMsg(*cloud_cluster[i], outputMsg);
// cout<<"data = "<< outputMsg <<endl;
cloud_cluster[i]->header.frame_id = FRAME_ID;
pclpub[i++].publish(outputMsg);
// i++;
}
}
else
ROS_INFO_STREAM("0 clusters extracted\n");
}
And this one is the main
int main(int argc, char** argv){
for (int z = 0; z < 10; z++) {
// std::cout << " - clustering/" << z << std::endl;
cloud_cluster[z] = pcl::PointCloud<pcl::PointXYZRGB>::Ptr(new pcl::PointCloud<pcl::PointXYZRGB>);
cloud_cluster[z]->height = 1;
cloud_cluster[z]->is_dense = true;
// cloud_cluster[z]->header.frame_id = FRAME_ID;
}
ros::init(argc,argv,"clustering");
ros::NodeHandlePtr nh(new ros::NodeHandle());
pclsub = nh->subscribe("/pclsegmen",1,cloudReceive);
std::string pub_str("clustering/0");
for (int z = 0; z < 10; z++) {
pub_str[11] = z + 48;//48=0(ASCII)
// z++;
pclpub[z] = nh->advertise <sensor_msgs::PointCloud2> (pub_str, 1);
}
// pclpub = nh->advertise<sensor_msgs::PointCloud2>("/pclcluster",1);
ros::spin();
}

This isn't an exact answer, but I think it addresses your issue & may ease your debugging.
RViz can directly subscribe to a published point cloud, the one I'm assuming you're trying to see in the cloud_receive callback. If you set the Frame to whichever frame it's being published at, and add it from the available topics, you should see the points. (Easier than trying to rebroadcast it as different topics).
Also, I recommend looking at the rostopic command line tool. You can do rostopic list to check if it's being published, rostopic bw to see if it's really publishing the expected volume of data (ex bytes vs kilobytes vs megabytes), rostopic hz to see how frequently (if ever) it's publishing, and (briefly) rostopic echo to look at the data itself. (This is me assuming from your question it's more an issue with the data coming into your node).
If you're having trouble, not with data coming into the node, nor with the visualization of pointcloud data in general, but with the transformed data that's supposed to come out of the node, I would check that the clustering worked, & reduce your code moreso to just having 1 publisher publish something. You may be doing something weird. Like messing up your pointers. You could also turn on stronger compilation warnings for your node with -Wall -Wextra -Werror or step through the execution of it via gdb (launch-prefix="xterm -e gdb --args").

The solution is, i change the ASCII number into lexical_cast. Thanks for your response, i hope this can help other
for (int z = 0; z < CLOUD_QTD; z++) {
// pub_str[11] = z + 48;
std::string topicName = "/pclcluster/" + boost::lexical_cast<std::string>(z);
global::pub[z] = n.advertise <sensor_msgs::PointCloud2> (topicName, 1);
}

Related

Visualizing a line in drake visualizer with C++

The question is related to Is there a way of visualising a line in drake visualizer where I had asked about how to visualize a line in the drake visualizer (about 3 years ago, which worked fine with v0.10.0). I am trying to achieve the same with the new API and was wondering if there was any example/documentation which can guide me on how to publish a line onto the visualizer. My previous method used for publishing a line looks like:
void publishLine(const std::vector<std::vector<double>>& pts,
const std::vector<std::string>& path, lcm::DrakeLcm& lcm,
std::vector<double> color) {
long long int now = getUnixTime() * 1000 * 1000;
nlohmann::json j = {{"timestamp", now},
{
"setgeometry",
{{{"path", path},
{"geometry",
{
{"type", "line"},
{"points", pts},
{"color", color},
{"radius", 0.1},
}}}},
},
{"settransform", nlohmann::json({})},
{"delete", nlohmann::json({})}};
auto msg = robotlocomotion::viewer2_comms_t();
msg.utime = now;
msg.format = "treeviewer_json";
msg.format_version_major = 1;
msg.format_version_minor = 0;
msg.data.clear();
for (auto& c : j.dump()) msg.data.push_back(c);
msg.num_bytes = j.dump().size();
// Use channel 0 for remote viewer communications.
lcm.get_lcm_instance()->publish("DIRECTOR_TREE_VIEWER_REQUEST_<0>", &msg);
}
You can use Meshcat::SetLine or Meshcat::SetLineSegments https://drake.mit.edu/doxygen_cxx/classdrake_1_1geometry_1_1_meshcat.html#aa5b082d79e267c040cbd066a11cdcb54
One caveat is that many browsers/webGL implementations do not support the linewidth property in ThreeJS. For thick lines, consider adding a cylinder using SetObject.

Understanding StereoMatching in Point Cloud Library

Situation: I am trying to get point cloud with pcl::AdaptiveCostSOStereoMatching, which uses two rectified images (pics are ok).
I used these tutorials to learn how to do this:
First tutorial
Second tutorial
Error: programm crashes in runtime when calling "compute" method of AdaptiveCostSOStereoMatching
Question: how to correctly pass images to "compute" method?
I tried:
1) Images converted by png2pcd
(command line: "png2pcd.exe in.png out.pcd")
2) Images converted with function below from cv::Mat
But no luck.
Function which converts cv::Mat to pcl::PointCloud
void MatToPointCloud(Mat& mat, pcl::PointCloud<RGB>::Ptr cloud)
{
int width = mat.cols;
int height = mat.rows;
pcl::RGB val;
val.r = 0; val.g = 0; val.b = 0;
for (int i = 0; i < mat.rows; i++)
for (int j = 0; j < mat.cols; j++)
{
auto point = mat.at<Vec3b>(i, j);
//std::cout << j << " " << i << "\n";
val.b = point[0];
val.g = point[1];
val.r = point[2];
cloud->at(j, i) = val;
}
}
pcl::AdaptiveCostSOStereoMatching (compute)
// Input
Mat leftMat, rightMat;
leftMat = imread("left.png");
rightMat = imread("right.png");
int width = leftMat.cols;
int height = rightMat.rows;
pcl::RGB val;
val.r = 0; val.g = 0; val.b = 0;
pcl::PointCloud<pcl::RGB>::Ptr left_cloud(new pcl::PointCloud<pcl::RGB>(width, height, val));
pcl::PointCloud<pcl::RGB>::Ptr right_cloud(new pcl::PointCloud<pcl::RGB>(width, height, val));
MatToPointCloud(leftMat, left_cloud);
MatToPointCloud(rightMat, right_cloud);
// Calculation
pcl::AdaptiveCostSOStereoMatching stereo;
stereo.setMaxDisparity(60);
//stereo.setXOffest(0); Почему-то не распознается
stereo.setRadius(5);
stereo.setSmoothWeak(20);
stereo.setSmoothStrong(100);
stereo.setGammaC(25);
stereo.setGammaS(10);
stereo.setRatioFilter(20);
stereo.setPeakFilter(0);
stereo.setLeftRightCheck(true);
stereo.setLeftRightCheckThreshold(1);
stereo.setPreProcessing(true);
stereo.compute(*left_cloud, *right_cloud); // <-- CRASHING THERE
stereo.medianFilter(4);
pcl::PointCloud<pcl::PointXYZRGB>::Ptr out_cloud(new pcl::PointCloud<pcl::PointXYZRGB>);
stereo.getPointCloud(318.11220, 224.334900, 368.534700, 0.8387445, out_cloud, left_cloud);
Error information:
Output log: HEAP[App.exe]:
Heap block at 0000006B0F828460 modified at 0000006B0F8284A8 past requested size of 38
App.exe has triggered a breakpoint.
left_cloud (a right cloud looks like left_cloud)
Mini question: if AdaptiveCostSOStereoMatching really allows build point cloud from 2 images, how ACSSM doing this without insintric and excentic parameters?
Problem: I downloaded and installed old version of PCL without stereo.
After that, I downloaded stereo from other PCL pack and add this library to my PCL pack. And it worked incorrectly.
Solution: I compilled PCL 1.8 and my programm is ok now.
OS: Windows
IDE: MSVS 12 2013 x64
If you will try to compile PCL, these links can help you:
Official-tutorial-1
Official-tutorial-2
Good help with FLANN and VTK
Example to verify installation

Getting vector which holding certain point

I've a vector of points from a grey section of an image and written like this:
std::vector<Point> vectorg;
for(i = 0; i <= hei - 1; i++) {
for(j = 0; j <= wid - 1; j++) {
if(mask(i,j) == 128) {
vectorg.push_back(Point(j,i));
}
}
}
Knowing what coordinates stored in certain cell is possible by:
cout << vectorg[0].x;
cout << vectorg[0].y;
The question is now the other way around, is it possible to know which cell holds certain coordinates?
Thanks a lot, I'm new here also with opencv programming, I'll be in your care.
Just do the following:
#include <algorithm>
// ...
Point p(searchedX, searchedY);
std::vector<Point>::iterator element = std::find(vectorg.begin(), vectorg.end(), p);
if (element != vectorg.end()) {
cout << (*element).x << endl;
cout << (*element).y << endl;
} else {
cout << "The point is not in the vector" << endl;
}
It may be overkill, but a way to do it (without doing a greedy exhaustive search) would be to build a FLANN index that will store the position of your points.
The feature matrix is made of the coordinates of your points.
Since OpenCV knows how to convert a vector to a matrix, you should be able to use your current vector as is.
Then, if you want only one point, just ask for the 1 nearest neighbour in the query (k parameter).
The bonus is, if you decide later that you need to have also the closest points in the neighborhood, just raise the value of k.
Sorry for the late response, and thanks for the answers, they inspired me indirectly.
I found an easy way to do it. This is by making another Mat which holds the numbers where the coordinates were saved.
std::vector<Point> vectorg;
cv::Mat_<int> Index =Mat_<int>::zeros(hei,wid);
for(i = 0; i <= hei - 1; i++) {
for(j = 0; j <= wid - 1; j++) {
if(mask(i,j) == 128) {
vectorg.push_back(Point(j,i));
Index(vector[count])=count;
count++;
}
}
}
This way I can know which cell holds certain coordinates by simply:
cout<<Index(36,362); //as example
Thanks a lot, I'll be in your care next time.

how do i decode, change, then re-encode a CORBA IOR file (Visibroker) in my Java client code?

I am writing code to ingest the IOR file generated by the team responsible for the server and use it to bind my client to their object. Sounds easy, right?
For some reason a bit beyond my grasp (having to do with firewalls, DMZs, etc.), the value for the server inside the IOR file is not something we can use. We have to modify it. However, the IOR string is encoded.
What does Visibroker provide that will let me decode the IOR string, change one or more values, then re-encode it and continue on as normal?
I've already looked into IORInterceptors and URL Naming but I don't think either will do the trick.
Thanks in advance!
When you feel like you need to hack an IOR, resist the urge to do so by writing code and whatnot to mangle it to your liking. IORs are meant to be created and dictated by the server that contains the referenced objects, so the moment you start mucking around in there, you're kinda "voiding your warranty".
Instead, spend your time finding the right way to make the IOR usable in your environment by having the server use an alternative hostname when it generates them. Most ORBs offer such a feature. I don't know Visibroker's particular configuration options at all, but a quick Google search revealed this page that shows a promising value:
vbroker.se.iiop_ts.host
Specifies the host name used by this server engine.
The default value, null, means use the host name from the system.
Hope that helps.
Long time ago I wrote IorParser for GNU Classpath, the code is available. It is a normal parser written being aware about the format, should not "void a warranty" I think. IOR contains multiple tagged profiles that are encapsulated very much like XML so we could parse/modify profiles that we need and understand and leave the rest untouched.
The profile we need to parse is TAG_INTERNET_IOP. It contains version number, host, port and object key. Code that reads and writes this profile can be found in gnu.IOR class. I am sorry this is part of the system library and not a nice piece of code to copy paste here but it should not be very difficult to rip it out with a couple of dependent classes.
This question has been repeatedly asked as CORBA :: Get the client ORB address and port with use of IIOP
Use the FixIOR tool (binary) from jacORB to patch the address and port of an IOR. Download the binary (unzip it) and run:
fixior <new-address> <new-port> <ior-file>
The tool will override the content of the IOR file with the 'patched' IOR
You can use IOR Parser to check the resulting IOR and compare it to your original IOR
Use this function to change the IOR. pass stringified IOR as first argument.
void hackIOR(const char* str, char* newIOR )
{
size_t s = (str ? strlen(str) : 0);
char temp[1000];
strcpy(newIOR,"IOR:");
const char *p = str;
s = (s-4)/2; // how many octets are there in the string
p += 4;
int i;
for (i=0; i<(int)s; i++) {
int j = i*2;
char v=0;
if (p[j] >= '0' && p[j] <= '9') {
v = ((p[j] - '0') << 4);
}
else if (p[j] >= 'a' && p[j] <= 'f') {
v = ((p[j] - 'a' + 10) << 4);
}
else if (p[j] >= 'A' && p[j] <= 'F') {
v = ((p[j] - 'A' + 10) << 4);
}
else
cout <<"invalid octet"<<endl;
if (p[j+1] >= '0' && p[j+1] <= '9') {
v += (p[j+1] - '0');
}
else if (p[j+1] >= 'a' && p[j+1] <= 'f') {
v += (p[j+1] - 'a' + 10);
}
else if (p[j+1] >= 'A' && p[j+1] <= 'F') {
v += (p[j+1] - 'A' + 10);
}
else
cout <<"invalid octet"<<endl;
temp[i]=v;
}
temp[i] = 0;
// Now temp has decoded IOR string. print it.
// Replace the object ID in temp.
// Encoded it back, with following code.
int temp1,temp2;
int l,k;
for(k = 0, l = 4 ; k < s ; k++)
{
temp1=temp2=temp[k];
temp1 &= 0x0F;
temp2 = temp2 & 0xF0;
temp2 = temp2 >> 4;
if(temp2 >=0 && temp2 <=9)
{
newIOR[l++] = temp2+'0';
}
else if(temp2 >=10 && temp2 <=15)
{
newIOR[l++] = temp2+'A'-10;
}
if(temp1 >=0 && temp1 <=9)
{
newIOR[l++] = temp1+'0';
}
else if(temp1 >=10 && temp1 <=15)
{
newIOR[l++] = temp1+'A'-10;
}
}
newIOR[l] = 0;
//new IOR is present in new variable newIOR.
}
Hope this works for you.

Checking if removing an edge in a graph will result in the graph splitting

I have a graph structure where I am removing edges one by one until some conditions are met. My brain has totally stopped and i can't find an efficient way to detect if removing an edge will result in my graph splitting in two or more graphs.
The bruteforce solution would be to do an bfs until one can reach all the nodes from a random node, but that will take too much time with large graphs...
Any ideas?
Edit: After a bit of search it seems what I am trying to do is very similar to the fleury's algorithm, where I need to find if an edge is a "bridge" or not.
Edges that make a graph disconnected when removed are called 'bridges'. You can find them in O(|V|+|E|) with a single depth-first search over the whole graph. A related algorithm finds all 'articulation points' (nodes that, if removed, makes the graph disconnected) follows. Any edge between two articulation-points is a bridge (you can test that in a second pass over all edges).
//
// g: graph; v: current vertex id;
// r_p: parents (r/w); r_a: ascents (r/w); r_ap: art. points, bool array (r/w)
// n_v: bfs order-of-visit
//
void dfs_art_i(graph *g, int v, int *r_p, int *r_v, int *r_a, int *r_ap, int *n_v) {
int i;
r_v[v] = *n_v;
r_a[v] = *n_v;
(*n_v) ++;
// printf("entering %d (nv = %d)\n", v, *n_v);
for (i=0; i<g->vertices[v].n_edges; i++) {
int w = g->vertices[v].edges[i].target;
// printf("\t evaluating %d->%d: ", v, w);
if (r_v[w] == -1) {
// printf("...\n");
// This is the first time we find this vertex
r_p[w] = v;
dfs_art_i(g, w, r_p, r_v, r_a, r_ap, n_v);
// printf("\n\t ... back in %d->%d", v, w);
if (r_a[w] >= r_v[v]) {
// printf(" - a[%d] %d >= v[%d] %d", w, r_a[w], v, r_v[v]);
// Articulation point found
r_ap[i] = 1;
}
if (r_a[w] < r_a[v]) {
// printf(" - a[%d] %d < a[%d] %d", w, r_a[w], v, r_a[v]);
r_a[v] = r_a[w];
}
// printf("\n");
}
else {
// printf("back");
// We have already found this vertex before
if (r_v[w] < r_a[v]) {
// printf(" - updating ascent to %d", r_v[w]);
r_a[v] = r_v[w];
}
// printf("\n");
}
}
}
int dfs_art(graph *g, int root, int *r_p, int *r_v, int *r_a, int *r_ap) {
int i, n_visited = 0, n_root_children = 0;
for (i=0; i<g->n_vertices; i++) {
r_p[i] = r_v[i] = r_a[i] = -1;
r_ap[i] = 0;
}
dfs_art_i(g, root, r_p, r_v, r_a, r_ap, &n_visitados);
// the root can only be an AP if it has more than 1 child
for (i=0; i<g->n_vertices; i++) {
if (r_p[i] == root) {
n_root_children ++;
}
}
r_ap[root] = n_root_children > 1 ? 1 : 0;
return 1;
}
If you remove the link between vertices A and B, can't you just check that you can still reach A from B after the edge removal? That's a little better than getting to all nodes from a random node.
How do you choose the edges to be removed?
Can you tell more about your problem domain?
Just how large Is your graph? maybe BFS is just fine!
After you wrote that you are trying to find out whether an edge is a bridge or not, I suggest
you remove edges in decreasing order of their betweenness measure.
Essentially, betweenness is a measure of an edges (or vertices) centrality in a graph.
Edges with higher value of betweenness have greater potential of being a bridge in a graph.
Look it up on the web, the algorithm is called 'Girvan-Newman algorithm'.

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