LineChart FX-删除实线 [英] LineChart FX - Delete solid line

问题描述

我对LineChart JavaFX图形有一个好奇的问题.

I have a curious question about the graph LineChart JavaFX.

我有这张图: 点在X轴上形成跳跃"(如我所得分的两个红色点所示)，因此JavaFX绘制了这两个点之间的线.如何删除每个跳转"之间的线?

I have this graph: dots forming a "jump" on the X axis (as shown by the two red points I scored) and therefore JavaFX draws me the line between these two points. How do I remove that line between each "jump"?

我发布代码:

public class ControllerIndividua {

    public static void plotIndividuaFull(String path, Stage stage, String name) {
        final NumberAxis xAxisIntensity = new NumberAxis(); //Dichiarazione asseX
        final NumberAxis yAxisIntensity = new NumberAxis();//Dichiarazione asseY

        DetectionS1.countS1();

        //Dichiarazione del tipo di grafico
        final LineChart<Number, Number> lineChartIntensity = new LineChart<Number, Number>(xAxisIntensity,yAxisIntensity);

        ArrayList<Double> extractedData; //Lista dei valori dello dell' intensità
        ArrayList<Double> extractedTime; //Lista del tempo
        ArrayList<Double> extractedS1; //Lista del tempo
        ArrayList<Double> extractedS1Time; //Lista del tempo

        //Gestione e settaggio del grafico
        lineChartIntensity.getData().clear();

        try {
            //Popolamento delle liste
            extractedTime = IntensityExtractor.pointsTime();
            extractedData = IntensityExtractor.pointsIntensity();
            extractedS1 = DetectionS1.S1pitch();
            extractedS1Time = DetectionS1.pointsS1Time();
            XYChart.Series<Number, Number> series = new XYChart.Series<Number, Number>();
            XYChart.Series<Number, Number> seriesS1 = new XYChart.Series<Number, Number>(); //Creazione seconda serie
            series.setName("Intensità di:\t" + name.toUpperCase());

            for (int j = 0; j < extractedS1.size(); j++) {
                seriesS1.getData().add(new XYChart.Data<Number, Number>(extractedS1Time.get(j), extractedS1.get(j)));
                lineChartIntensity.getStyleClass().add("CSSintensity");
            }

            //Creazione finestra e stampa del grafico
            Scene scene = new Scene(lineChartIntensity, 1000, 600);
            lineChartIntensity.getData().addAll(series,seriesS1);
            scene.getStylesheets().add("application/application.css");
            stage.setScene(scene);
            stage.show();
        } catch (java.lang.Exception e) {
            e.printStackTrace();
        }
    }
}

有人对我该怎么做还有些想法?

Someone also has a little idea on how I could do?

谢谢大家.

推荐答案

这是一个古老的问题，答案已被接受，但我碰到了这个问题，对此感到好奇.我想知道是否有可能在LineChart中加一个空格(至少不必创建自定义图表实现).原来有.解决方案有点笨拙.它涉及获取 Path 通过使用 XYChart.Series.getNode() 并处理 PathElement s.下面的代码给出了一个示例:

This is an old question with an accepted answer but I came across it and was curious. I wanted to know if it was possible to put a gap in a LineChart (at least without having to create a custom chart implementation). It turns out that there is. The solution is kind of hacky and brittle. It involves getting the Path by using XYChart.Series.getNode() and manipulating the list of PathElements. The following code gives an example:

import javafx.application.Application;
import javafx.application.Platform;
import javafx.scene.Scene;
import javafx.scene.chart.LineChart;
import javafx.scene.chart.NumberAxis;
import javafx.scene.chart.XYChart;
import javafx.scene.layout.StackPane;
import javafx.scene.shape.LineTo;
import javafx.scene.shape.MoveTo;
import javafx.scene.shape.Path;
import javafx.stage.Stage;

public class Main extends Application {

    @Override
    public void start(Stage primaryStage) {
        LineChart<Number, Number> chart = new LineChart<>(new NumberAxis(), new NumberAxis());
        chart.getXAxis().setLabel("X");
        chart.getYAxis().setLabel("Y");
        chart.setLegendVisible(false);
        chart.getData().add(new XYChart.Series<>());

        for (int x = 0; x <= 10; x++) {
            chart.getData().get(0).getData().add(new XYChart.Data<>(x, Math.pow(x, 2)));
        }

        /*
         * Had to wrap the call in a Platform.runLater otherwise the Path was
         * redrawn after the modifications are made.
         */
        primaryStage.setOnShown(we -> Platform.runLater(() -> {
            Path path = (Path) chart.getData().get(0).getNode();
            LineTo lineTo = (LineTo) path.getElements().get(8);
            path.getElements().set(8, new MoveTo(lineTo.getX(), lineTo.getY()));
        }));

        primaryStage.setScene(new Scene(new StackPane(chart), 500, 300));
        primaryStage.setTitle("LineChart Gap");
        primaryStage.show();
    }

}

此代码将导致以下结果:

This code results in the following:

这是可能的，因为PathElement的ObservableList似乎是

This is possible because the ObservableList of PathElements seems to be a MoveTo followed by a bunch of LineTos. I simply picked a LineTo and replaced it with a MoveTo to the same coordinates. I haven't quite figured out which index of LineTo matches with which XYChart.Data, however, and picked 8 for the example randomly.

此解决方案有两个问题.第一个显而易见的是，这依赖于LineChart的内部实现.第二个是真正的脆性来自何处.数据的任何更改(轴的值范围，图表的宽度或高度)或导致图表重新绘制的几乎所有 都会导致重新计算Path并重新绘制.这意味着，如果您使用此解决方案，则必须每次 重新应用修改，图表会重新绘制.

There are a couple of issues with this solution. The first and obvious one is that this relies on the internal implementation of LineChart. The second is where the real brittleness comes from though. Any change to the data, either axes' value ranges, the chart's width or height, or pretty much anything that causes the chart to redraw itself will cause the Path to be recomputed and redrawn. This means if you use this solution you'll have to reapply the modification every time the chart redraws itself.

这篇关于LineChart FX-删除实线的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！