我有一个由0和1组成的矩阵。我想像网格一样可视化此矩阵,并在其中有1的单元格上标记。我进行了研究,以便找到在Java FX中执行此操作的方法,但找不到与我说的内容类似的内容。在Java FX中有什么方法可以做到这一点?最好的问候,
答案 0 :(得分:1)
我在Java FX中使用GridPane和Timeline可视化了矩阵。由于上面c0der的评论,我找到了这种方法。在我的情况下,我需要在矩阵上绘制一条路径,就像找到最小路径问题一样。这是我的源代码:
public class VisualizationThread implements EventHandler<ActionEvent> {
private ArrayList<ArrayList<Integer>> matrix; // Matrix
private ArrayList<Integer> solution; // One individual
private ArrayList<ArrayList<Integer>> visited; // Visited coordinate list
private int currentX; // Current x coordinate
private int currentY; // Current y coordinate
private int index; // Current movement index
private int attempt; // unsuccessful attempt count (used while visualizing all generations)
private boolean closeAtTheAnd; // Close all stages except the last one which shows final situation, the correct path
private boolean isSuccessful; // Currently visualizing an unsuccessful attempt or correct path?
private Stage primaryStage; // primary stage (gui component)
public VisualizationThread(ArrayList<ArrayList<Integer>> matrix, ArrayList<Integer> solution, int currentX, int currentY,
boolean isSuccessful, int attempt, boolean closeAtTheAnd){
this.matrix = matrix;
this.solution = solution;
this.currentY = currentY;
this.currentX = currentX;
this.isSuccessful = isSuccessful;
this.attempt = attempt;
this.primaryStage = new Stage();
this.index = 0;
this.closeAtTheAnd = closeAtTheAnd;
this.visited = new ArrayList<ArrayList<Integer>>();
}
// Visualize a single individual
public void run() {
// GUI processes
BorderPane root = new BorderPane();
GridPane grid = new GridPane();
grid.setPadding(new Insets(10));
grid.setHgap(10);
grid.setVgap(10);
StackPane[][] screen_buttons = new StackPane[matrix.size()][matrix.size()];
visited.add(new ArrayList<Integer>(){{add(currentX); add(currentY);}});
for (int y=0;y<matrix.size();y++) {
for (int x=0;x<matrix.get(y).size();x++) {
screen_buttons[y][x] = new StackPane();
Rectangle rec = new Rectangle(60,60);
// Visualize visited points as green and others yellow and foods red
if(!(currentX>=matrix.size() || currentY>=matrix.size())){
if(isVisited(visited, x, y)){
rec.setFill(Color.GREEN);
}else {
rec.setFill(matrix.get(y).get(x) == 0 ? Color.YELLOW : Color.RED);
}
}else {
rec.setFill(matrix.get(y).get(x) == 0 ? Color.YELLOW : Color.RED);
}
rec.setStyle("-fx-arc-height: 20; -fx-arc-width: 20;");
screen_buttons[y][x].getChildren().addAll(rec);
grid.add(screen_buttons[y][x], x, y);
}
}
//container for controls
GridPane controls = new GridPane();
root.setCenter(grid);
root.setBottom(controls);
Scene scene = new Scene(root);
primaryStage.setScene(scene);
primaryStage.setTitle(isSuccessful ? "Correct Path" : "Unsuccessful Attempt: " + attempt);
primaryStage.show();
if(!(index >= solution.size())){
switch (solution.get(index)){
case 1:
setCurrentY(getCurrentY()-1);
break;
case 2:
setCurrentX(getCurrentX()-1);
break;
case 3:
setCurrentY(getCurrentY()+1);
break;
case 4:
setCurrentX(getCurrentX()+1);
break;
}
index++;
}
if(closeAtTheAnd && index == solution.size()){
primaryStage.close();
}
}
// This method finds if current coordinate is visited before or not
// @param visitedList: List of visited points
// @param x: x coordinate
// @param y: y coordinate
public boolean isVisited(ArrayList<ArrayList<Integer>> visitedList, int x, int y){
boolean result = false;
// Check if x and y coordinates are visited before
for (ArrayList<Integer> temp : visitedList){
if(temp.get(0) == x && temp.get(1) == y){
result = true;
break;
}
}
return result;
}
public int getCurrentX() {
return currentX;
}
public void setCurrentX(int currentX) {
this.currentX = currentX;
}
public int getCurrentY() {
return currentY;
}
public void setCurrentY(int currentY) {
this.currentY = currentY;
}
public void handle(ActionEvent event) {
run();
}
}
// In another part of the code which I need to trigger the event above.
ArrayList<Integer> bestOfLastPopulation = geneticAlgorithm.findMostAte(finalPopulation, matrix, centerX, centerY);
VisualizationThread visualizationThread = new VisualizationThread(matrix, bestOfLastPopulation, centerX, centerY, true, 0, false);
Timeline fiveSecondsWonder = new Timeline(new KeyFrame(Duration.millis(150), visualizationThread));
fiveSecondsWonder.setCycleCount(bestOfLastPopulation.size() + 1);
fiveSecondsWonder.play();
对于试图找到相同事物的人,应查看以嵌套的for循环开始并以primaryStage.show()结尾的部分。在代码的这一部分,它以不同的颜色可视化矩阵。例如,如果一个单元格包含1,则为红色,如果一个单元格包含0,则为黄色,如果之前访问过该单元格,则将其涂成绿色。希望对别人有帮助。最好的问候,