I want to use Retrofit w/ RxJava to make API requests for items that are a certain distance from a given location. For illustation, this is how it might it may look for Retrofit w/o RxJava:
API api = ...
List<Item> items = new ArrayList<Item>();
int miles = 50;
do {
items = api.getItems(lon, lat, distance);
miles += 50;
} while (items.size() == 0);
Essentially, we'll keep increasing the distance until we get a response that has at least 1 item.
What's the best way to handle this kind of workflow with RxJava?
You can a combination of range, concatMap and filter for this purpose:
static Observable<Integer> getItems(int distance) {
if (distance < 500) {
return Observable.<Integer>empty().delay(500, TimeUnit.MILLISECONDS);
}
return Observable.just(1).delay(500, TimeUnit.MILLISECONDS);
}
public static void main(String[] args) {
Observable
.range(1, 20)
.map(v -> 50 * v)
.concatMap(d -> getItems(d).toList())
.doOnNext(list -> System.out.println("Got a list of " + list.size() + " items"))
.filter(list -> !list.isEmpty())
.first()
.toBlocking()
.forEach(System.out::println);
;
}
Related
I use ListChangeListener to listen to changes in Tab Pane.
private final TabPane tabBar = new TabPane();
...
tabBar.getTabs().addListener(this::tabsChanged);
I'm trying to listen to tab move events with the following code:
private void tabsChanged(ListChangeListener.Change<? extends Tab> change) {
while (change.next()) {
if (change.wasPermutated()) {
for (int i = change.getFrom(); i < change.getTo(); i++) {
System.out.println(i + " -> " + change.getPermutation(i));
}
}
}
}
As JavaFX documentation says:
In order to get the new position of an element, you must call:
change.getPermutation(oldIndex). Returns: the new index of the same
element.
But in my case change.getPermutation(i) always returns just i.
For example, I have 4 tabs.
Their indexes are: 0, 1, 2, 3.
Then I move the 4th tab to the first position.
I expect the following output:
0 -> 1
1 -> 2
2 -> 3
3 -> 0
But I get:
0 -> 0
1 -> 1
2 -> 2
3 -> 3
How can I make it work as I need?
As already noted in the comments: the behavior you observe is a bug just reported as JDK-8278062 - the doc and your expectation based on the doc is correct, the notification (implemented in the internal class TabObservableList) is wrong.
Normally, if we want to find the newIndex, a listChangeListener would do something like:
for (int oldIndex = c.getFrom(); oldIndex < c.getTo(); ++oldIndex) {
int newIndex = c.getPermutation(oldIndex);
...
}
To work around the issue, we could manually keep a copy of the tabs, lookup the tab at the old index and find its new index in the re-ordered tabs:
for (int oldIndex = c.getFrom(); oldIndex < c.getTo(); ++oldIndex) {
Tab tab = copy.get(oldIndex);
int newIndex = c.getList().indexOf(tab);
...
}
// update the copy
Or we could have some fun and implement a TransformationList around the original tabs that does the work for us :) It jumps in when it detects a permutation and fires the correct notification. Note that the only internal class used below is SourceChangeAdapter, we either need to relax encapsulation or c&p its content (it is doing nothing but pass on notifications on behalf of the wrapper)
public class TabObservableListWrapper extends TransformationList<Tab, Tab> {
// copy of source used to build the correct permutation
private ObservableList<Tab> copy = FXCollections.observableArrayList();
public TabObservableListWrapper(ObservableList<Tab> source) {
super(source);
updateCopy();
}
#Override
protected void sourceChanged(Change<? extends Tab> c) {
// TBD: cope with a change that has
// - a mixture of permutation and other subchanges
// - multiple subchanges of type permutation
boolean isPermutation = false;
// check if the change is a permutation
while (c.next()) {
if (c.wasPermutated()) {
isPermutation = true;
break;
}
}
c.reset();
if (isPermutation) {
beginChange();
updatePermutation(c);
endChange();
} else {
// assuming other change type notifications are correct, just delegate
fireChange(new SourceAdapterChange<>(this, c));
}
// keep copy sync'ed to source
updateCopy();
}
/**
* Converts the incorrect permutation notification from source
* into a correct one and let super fire the appropriate change.
*
* Note: this method must be called inside a begin/endChange block.
* #param c a change with a single subChange of type wasPermutated
*/
private void updatePermutation(Change<? extends Tab> c) {
c.next();
int from = c.getFrom();
int to = c.getTo();
int permSize = to - from;
int[] perm = new int[permSize];
// fill the perm
for(int i = 0; i < permSize; i++) {
int oldIndex = from + i;
Tab tab = copy.get(oldIndex);
perm[i] = c.getList().indexOf(tab);
}
nextPermutation(from, to, perm);
}
// keep copy sync'ed
private void updateCopy() {
copy.setAll(getSource());
}
// implement public methods by delegating 1:1 to source
#Override
public int getSourceIndex(int index) {
return index;
}
#Override
public int getViewIndex(int index) {
return index;
}
#Override
public Tab get(int index) {
return getSource().get(index);
}
#Override
public int size() {
return getSource().size();
}
}
To use, wrap it around a tabPane's tab list and listen to the wrapper instead of directly to original list, something like:
TabObservableListWrapper wrapper = new TabObservableListWrapper(tabPane.getTabs());
wrapper.addListener((ListChangeListener<Tab>)change -> {
while (change.next()) {
if (change.wasPermutated()) {
System.out.println("from wrapper:");
for (int oldIndex = change.getFrom(); oldIndex < change.getTo(); oldIndex++) {
System.out.println(oldIndex + " -> " + change.getPermutation(oldIndex));
}
}
}
});
I have an enormous directed graph I need to traverse in search for the shortest path to a specific node from a given starting point. The graph in question does not exist explicitly; the child nodes are determined algorithmically from the parent nodes.
(To give an illustration: imagine a graph of chess positions. Each node is a chess position and its children are all the legal moves from that position.)
So I have a queue for open nodes, and every time I process the next node in the queue I enqueue all of its children. But since the graph can have cycles I also need to maintain a hashset of all visited nodes so I can check if I have visited one before.
This works okay, but since this graph is so large, I run into memory problems. All of the nodes in the queue are also stored in the hashset, which tends to be around 50% of the total number or visited nodes in practice in my case.
Is there some magical way to get rid of this redundancy while keeping the speed of the hashset? (Obviously, I could get rid of the redundancy by NOT hashing and just doing a linear search, but that is out of the question.)
I solved it by writing a class that stores the keys in a list and stores the indices of the keys in a hashtable. The next node "in the queue" is always the the next node in the list until you find what you're looking for or you've traversed the entire graph.
class IndexMap<T>
{
private List<T> values;
private LinkedList<int>[] buckets;
public int Count { get; private set; } = 0;
public IndexMap(int capacity)
{
values = new List<T>(capacity);
buckets = new LinkedList<int>[NextPowerOfTwo(capacity)];
for (int i = 0; i < buckets.Length; ++i)
buckets[i] = new LinkedList<int>();
}
public void Add(T item) //assumes item is not yet in map
{
if (Count == buckets.Length)
ReHash();
int bucketIndex = item.GetHashCode() & (buckets.Length - 1);
buckets[bucketIndex].AddFirst(Count++);
values.Add(item);
}
public bool Contains(T item)
{
int bucketIndex = item.GetHashCode() & (buckets.Length - 1);
foreach(int i in buckets[bucketIndex])
{
if (values[i].Equals(item))
return true;
}
return false;
}
public T this[int index]
{
get => values[index];
}
private void ReHash()
{
LinkedList<int>[] newBuckets = new LinkedList<int>[2 * buckets.Length];
for (int i = 0; i < newBuckets.Length; ++i)
newBuckets[i] = new LinkedList<int>();
for (int i = 0; i < buckets.Length; ++i)
{
foreach (int index in buckets[i])
{
int bucketIndex = values[index].GetHashCode() & (newBuckets.Length - 1);
newBuckets[bucketIndex].AddFirst(index);
}
buckets[i] = null;
}
buckets = newBuckets;
}
private int NextPowerOfTwo(int n)
{
if ((n & n-1) == 0)
return n;
int output = 0;
while (n > output)
{
output <<= 1;
}
return output;
}
}
The old method of maintaining both an array of the open nodes and a hashtable of the visited nodes needed n*(1+a)*size(T) space, where a is the ratio of nodes_in_the_queue over total_nodes_found and size(T) is the size of a node.
This method needs n*(size(T) + size(int)). If your nodes are significantly larger than an int, this can save a lot.
Generalities : explanations about my program and its functioning
I am working on a photo-retouching JavaFX application. The final user can load several images. When he clicks on the button REVERSE, a Task is launched for each image using an Executor. Each of these Task executes the reversal algorithm : it fills an ArrayBlockingQueue<Pixel> (using add method).
When the final user clicks on the button REVERSE, as I said, these Task are launched. But just after these statements, I tell the JavaFX Application Thread to draw the Pixel of the ArrayBlockingQueue<Pixel> (using remove method).
Thus, there are parallelism and concurrency (solved by the ArrayBlockingQueue<Pixel>) between the JavaFX Application Thread and the Task, and between the Task themselves.
To draw the Pixel of the ArrayBlockingQueue<Pixel>, the JavaFX Application Thread starts an AnimationTimer. The latter contains the previously-mentionned remove method. This AnimationTimer is started for each image.
I think you're wondering yourself how this AnimationTimer can know to what image belongs the Pixel it has removed ? In fact, each Pixel has an attribute writable_image that specifies the image to what it belongs.
My problems
Tell me if I'm wrong, but my program should work. Indeed :
My JavaFX Application Thread is the only thread that change the GUI (and it's required in JavaFX) : the Task just do the calculations.
There is not concurrency, thanks to the BlockingQueue I use (in particular, there isn't possibility of draining).
The AnimationTimer knows to what image belongs each Pixel.
However, it's (obviously !) not the case (otherwise I wouldn't have created this question haha !).
My problem is that my JavaFX Application freezes (first problem), after having drawn only some reversed pixels (not all the pixels). On the last loaded image moreover (third problem).
A detail that could be the problems' cause
But I would need your opinion.
The AnimationTimer of course doesn't draw the reversed pixels of each image directly : this is animated. The final user can see each pixel of an image being reversed, little by little. It's very practical in other algorithms as the creation of a circle, because the user can "look" how the algorithm works.
But to do that, the AnimationTimer needs to read a variable called max. This variable is modified (writen) in... each Task. But it's an AtomicLong. So IF I AM NOT WRONG, there isn't any problem of concurrency between the Task themselves, or between the JavaFX Application Thread and these Task.
However, it could be the problem : indeed, the max's value could be 2000 in Task n°1 (= in image n°1), and 59 in Task n°2 (= in image n°2). The problem is the AnimationTimer must use 2000 for the image n°1, and 59 for the n°2. But if the Task n°1 et n°2 have finished, the only value known by the AnimationTimer would be 59...
Sources
When the user clicks on the button REVERSE
We launch the several Task and start several times the AnimationTimer. CLASS : RightPane.java
WritableImage current_writable_image;
for(int i = 0; i < this.gui.getArrayListImageViewsImpacted().size(); i++) {
current_writable_image = (WritableImage) this.gui.getArrayListImageViewsImpacted().get(i).getImage();
this.gui.getGraphicEngine().executor.execute(this.gui.getGraphicEngine().createTask(current_writable_image));
}
for(int i = 0; i < this.gui.getArrayListImageViewsImpacted().size(); i++) {
current_writable_image = (WritableImage) this.gui.getArrayListImageViewsImpacted().get(i).getImage();
this.gui.getImageAnimation().setWritableImage(current_writable_image);
this.gui.getImageAnimation().startAnimation();
}
The Task are part of the CLASS GraphicEngine, which contains an Executor :
public final Executor executor = Executors.newCachedThreadPool(runnable -> {
Thread t = new Thread(runnable);
t.setDaemon(true);
return t ;
});
public Task createTask(WritableImage writable_image) {
int image_width = (int) writable_image.getWidth(), image_height = (int) writable_image.getHeight();
Task ret = new Task() {
protected Void call() {
switch(operation_to_do) {
case "reverse" :
gui.getImageAnimation().setMax(image_width*image_height); // USE OF "MAX" VARIABLE
reverseImg(writable_image);
break;
}
return null;
}
};
return ret;
}
The same CLASS, GraphicEngine, also contains the reversal algorithm :
private void reverseImg(WritableImage writable_image) {
int image_width = (int) writable_image.getWidth(), image_height = (int) writable_image.getHeight();
BlockingQueue<Pixel> updates = gui.getUpdates();
PixelReader pixel_reader = writable_image.getPixelReader();
double[] rgb_reversed;
for (int x = 0; x < image_width; x++) {
for (int y = 0; y < image_height; y++) {
rgb_reversed = PhotoRetouchingFormulas.reverse(pixel_reader.getColor(x, y).getRed(), pixel_reader.getColor(x, y).getGreen(), pixel_reader.getColor(x, y).getBlue());
updates.add(new Pixel(x, y, Color.color(rgb_reversed[0], rgb_reversed[1], rgb_reversed[2], pixel_reader.getColor(x, y).getOpacity()), writable_image));
}
}
}
Finally, here is the code of the CLASS AnimationTimer. There is nothing particular. Note the variable max is used here too (and in the CLASS GraphicEngine : setMax).
public class ImageAnimation extends AnimationTimer {
private Gui gui;
private AtomicLong max, speed, max_delay;
private long count, start;
private WritableImage writable_image;
ImageAnimation (Gui gui) {
this.gui = gui;
this.count = 0;
this.start = -1;
this.max = new AtomicLong(Long.MAX_VALUE);
this.max_delay = new AtomicLong(999_000_000);
this.speed = new AtomicLong(this.max_delay.get());
}
public void setMax(long max) {
this.max.set(max);
}
public void setSpeed(long speed) { this.speed.set(speed); }
public double getMaxDelay() { return this.max_delay.get(); }
#Override
public void handle(long timestamp) {
if (start < 0) {
start = timestamp ;
return ;
}
ArrayList<Pixel> list_sorted_pixels = new ArrayList<>();
BlockingQueue<Pixel> updates = this.gui.getUpdates();
for(Pixel new_pixel : updates) {
if(new_pixel.getWritableImage() == writable_image) {
list_sorted_pixels.add(new_pixel);
}
}
while (list_sorted_pixels.size() > 0 && timestamp - start > (count * this.speed.get()) / (writable_image.getWidth()) && !updates.isEmpty()) {
Pixel update = list_sorted_pixels.remove(0);
updates.remove(update);
count++;
if (update.getX() >= 0 && update.getY() >= 0) {
writable_image.getPixelWriter().setColor(update.getX(), update.getY(), update.getColor());
}
}
if (count >= max.get()) {
this.count = 0;
this.start = -1;
this.max.set(Long.MAX_VALUE);
stop();
}
}
public void setWritableImage(WritableImage writable_image) { this.writable_image = writable_image; }
public void startAnimation() {
this.start();
}
}
Given a directed graph with
A root node
Some leaves nodes
Multiple nodes can be connected to the same node
Cycles can exist
We need to print all the paths from the root node to all the leaves nodes. This is the closest question I got to this problem
Find all paths between two graph nodes
If you actually care about ordering your paths from shortest path to longest path then it would be far better to use a modified A* or Dijkstra Algorithm. With a slight modification the algorithm will return as many of the possible paths as you want in order of shortest path first. So if what you really want are all possible paths ordered from shortest to longest then this is the way to go. The code I suggested above would be much slower than it needs to be if you care about ordering from shortest to longest, not to mention would take up more space then you'd want in order to store every possible path at once.
If you want an A* based implementation capable of returning all paths ordered from the shortest to the longest, the following will accomplish that. It has several advantages. First off it is efficient at sorting from shortest to longest. Also it computes each additional path only when needed, so if you stop early because you dont need every single path you save some processing time. It also reuses data for subsequent paths each time it calculates the next path so it is more efficient. Finally if you find some desired path you can abort early saving some computation time. Overall this should be the most efficient algorithm if you care about sorting by path length.
import java.util.*;
public class AstarSearch {
private final Map<Integer, Set<Neighbor>> adjacency;
private final int destination;
private final NavigableSet<Step> pending = new TreeSet<>();
public AstarSearch(Map<Integer, Set<Neighbor>> adjacency, int source, int destination) {
this.adjacency = adjacency;
this.destination = destination;
this.pending.add(new Step(source, null, 0));
}
public List<Integer> nextShortestPath() {
Step current = this.pending.pollFirst();
while( current != null) {
if( current.getId() == this.destination )
return current.generatePath();
for (Neighbor neighbor : this.adjacency.get(current.id)) {
if(!current.seen(neighbor.getId())) {
final Step nextStep = new Step(neighbor.getId(), current, current.cost + neighbor.cost + predictCost(neighbor.id, this.destination));
this.pending.add(nextStep);
}
}
current = this.pending.pollFirst();
}
return null;
}
protected int predictCost(int source, int destination) {
return 0; //Behaves identical to Dijkstra's algorithm, override to make it A*
}
private static class Step implements Comparable<Step> {
final int id;
final Step parent;
final int cost;
public Step(int id, Step parent, int cost) {
this.id = id;
this.parent = parent;
this.cost = cost;
}
public int getId() {
return id;
}
public Step getParent() {
return parent;
}
public int getCost() {
return cost;
}
public boolean seen(int node) {
if(this.id == node)
return true;
else if(parent == null)
return false;
else
return this.parent.seen(node);
}
public List<Integer> generatePath() {
final List<Integer> path;
if(this.parent != null)
path = this.parent.generatePath();
else
path = new ArrayList<>();
path.add(this.id);
return path;
}
#Override
public int compareTo(Step step) {
if(step == null)
return 1;
if( this.cost != step.cost)
return Integer.compare(this.cost, step.cost);
if( this.id != step.id )
return Integer.compare(this.id, step.id);
if( this.parent != null )
this.parent.compareTo(step.parent);
if(step.parent == null)
return 0;
return -1;
}
#Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
Step step = (Step) o;
return id == step.id &&
cost == step.cost &&
Objects.equals(parent, step.parent);
}
#Override
public int hashCode() {
return Objects.hash(id, parent, cost);
}
}
/*******************************************************
* Everything below here just sets up your adjacency *
* It will just be helpful for you to be able to test *
* It isnt part of the actual A* search algorithm *
********************************************************/
private static class Neighbor {
final int id;
final int cost;
public Neighbor(int id, int cost) {
this.id = id;
this.cost = cost;
}
public int getId() {
return id;
}
public int getCost() {
return cost;
}
}
public static void main(String[] args) {
final Map<Integer, Set<Neighbor>> adjacency = createAdjacency();
final AstarSearch search = new AstarSearch(adjacency, 1, 4);
System.out.println("printing all paths from shortest to longest...");
List<Integer> path = search.nextShortestPath();
while(path != null) {
System.out.println(path);
path = search.nextShortestPath();
}
}
private static Map<Integer, Set<Neighbor>> createAdjacency() {
final Map<Integer, Set<Neighbor>> adjacency = new HashMap<>();
//This sets up the adjacencies. In this case all adjacencies have a cost of 1, but they dont need to.
addAdjacency(adjacency, 1,2,1,5,1); //{1 | 2,5}
addAdjacency(adjacency, 2,1,1,3,1,4,1,5,1); //{2 | 1,3,4,5}
addAdjacency(adjacency, 3,2,1,5,1); //{3 | 2,5}
addAdjacency(adjacency, 4,2,1); //{4 | 2}
addAdjacency(adjacency, 5,1,1,2,1,3,1); //{5 | 1,2,3}
return Collections.unmodifiableMap(adjacency);
}
private static void addAdjacency(Map<Integer, Set<Neighbor>> adjacency, int source, Integer... dests) {
if( dests.length % 2 != 0)
throw new IllegalArgumentException("dests must have an equal number of arguments, each pair is the id and cost for that traversal");
final Set<Neighbor> destinations = new HashSet<>();
for(int i = 0; i < dests.length; i+=2)
destinations.add(new Neighbor(dests[i], dests[i+1]));
adjacency.put(source, Collections.unmodifiableSet(destinations));
}
}
The output from the above code is the following:
[1, 2, 4]
[1, 5, 2, 4]
[1, 5, 3, 2, 4]
Notice that each time you call nextShortestPath() it generates the next shortest path for you on demand. It only calculates the extra steps needed and doesnt traverse any old paths twice. Moreover if you decide you dont need all the paths and end execution early you've saved yourself considerable computation time. You only compute up to the number of paths you need and no more.
Finally it should be noted that the A* and Dijkstra algorithms do have some minor limitations, though I dont think it would effect you. Namely it will not work right on a graph that has negative weights.
Here is a link to JDoodle where you can run the code yourself in the browser and see it working. You can also change around the graph to show it works on other graphs as well: http://jdoodle.com/a/ukx
I ask IN NUTITEQ, I know the way to perform this with GoogleApi with computeArea() but I have not found anything in Nutiteq sdk/snapshot.
Thanks in advance.
p.s. I know many methods to compute the area but I want to call something own of Nutiteq
EDIT:
There are no built-in Method, thanks for the fast answer Jaak, so I researched and found 2 Methods, both of them under WSG84 Projection. I developed a little programm to compare both of them and then I compared to a KML Tool, which computes area of a Polygon.
how-can-i-measure-area-from-geographic-coordinates
A-link-to-a-Geographic-Framework
And the results with a kml tool of University of New Hampshire
12197.38184
import java.lang.Math.*;
import net.sf.geographiclib.*;
public class ComputeAreaTest {
private static double[][] moorwiese_coords = {{12.925634f,48.427168f},
{12.926825f,48.427217f},
{12.926788f,48.428385f},
{12.926069f,48.428374f},
{12.925431f,48.42825f},
{12.925624f,48.427192f},
{12.925634f,48.427168f}};
protected static double computeArea() {
double area=0.0;
double earthRadius = 6378137.0f;
int size = moorwiese_coords.length;
if (size > 2) {
double[] p1 = new double[2];
double[] p2 = new double[2];
for (int i=0; i<size-1; i++) {
p1 = moorwiese_coords[i];
p2 = moorwiese_coords[i+1];
area += Math.toRadians(p2[0] - p1[0]) * (2 +
Math.sin(Math.toRadians(p1[1]) ) + Math.sin(Math.toRadians(p2[1])) );
}
area = area * earthRadius * earthRadius / 2.0;
}
return area;
}
protected static double computeAreaWithGeographicLib() {
int size = moorwiese_coords.length;
PolygonArea p = new PolygonArea(Geodesic.WGS84, false);
try {
for (int i=0;i<size;i++) {
p.AddPoint(moorwiese_coords[i][4], moorwiese_coords[i][0]);
}
}
catch (Exception e) {}
PolygonResult r = p.Compute();
return r.area;
}
public static void main(String[] args) {
double areaGeoLib = computeAreaWithGeographicLib();
double area = computeArea();
System.out.println("Area: " + (-1)*area + "\nArea(GeoLib): "+areaGeoLib);
}
}
Output
Area: 12245.282587113787
Area(GeoLib): 12254.95547034964
I found not very suitable for accurate use ( Yes, under 0.5% Error may be unaccurate for many environments) but useful to learn how to compute the area of a irregular Polygon.
Thx, Jaak, I had not seen this freen Symbol to click "Right Answer", now I have seen, I must explicitly. So, an "answer" is in initial post.