[我要问的基本问题是:如何制作一个图,其中y=0
处的线实际上位于y=*some number*
,并且随着Y的增加,Java的y减少?
我正在尝试制作折线图来记录动物种群的变化。我需要围绕Y=0
线制作图表的y=980
。我还需要做一些会引起人口增加的事情,并用y的减少来表示它(使线上升)。我要说的是,我需要创建一个看起来像折线图的折线图。
我尝试了多种不同的方法,根据不同的输入,每种方法都会给我带来不同的结果。我已经成功创建了从y=980
线开始并向上移动的图形,如下图所示。我用于此的方法是绘制线条(用于图形本身),然后取差值的绝对值减去值减去90的10倍,如[]所示。
g.drawLine((125), (y+1)*93, width, (y+1)*93); g.drawString(Math.abs(y*10-90) + " alive", (80), (y+1)*93);
这完全适用于图形,但是当尝试将其实现到线的图形中时,会收到混合结果。
对于此项目,我有3个种群:我的动物的捕食者,我的动物的猎物和我的动物。我想绘制所有这些的人口规模。当使用前面显示的方法时,我成功地绘制了捕食者种群的第一个值。但是,其他两个位于图的相对位置。 (例如,我的动物的种群大小设置为从90开始,但是在图表上大约是20年代中期,也如下图所示)。我用于这些代码的代码是:
// predator animal line g.setColor(Color.red); g.drawLine(i*93+125, (Math.abs((predatorAnimalAmt[i]*10)-90)), (i+1)*93+125, (Math.abs((predatorAnimalAmt[i+1]*10)-90))); g.drawLine(100, height-45, 120, height-45); g.drawString("PREDATOR", 30, height-40); // prey animal line g.setColor(Color.green); g.drawLine(i*93+125, (Math.abs((preyAnimalAmt[i]*10)-90)), (i+1)*93+125, (Math.abs((preyAnimalAmt[i+1]*10)-90))); g.drawLine(100, height-60, 120, height-60); g.drawString("PREY", 30, height-55); // our animal's line g.setColor(Color.blue); g.drawLine(i*93+125, (Math.abs((ourAnimalAmt[i]*10)-90)), (i+1)*93+125, (Math.abs((ourAnimalAmt[i+1]*10)-90))); g.drawLine(100, height-75, 120, height-75); g.drawString("OUR ANIMAL", 30, height-70);
这是我在此类中的代码(my github page中还有其他可访问的类)
import java.awt.Color; import java.awt.Graphics; import java.awt.Toolkit; import java.awt.image.BufferStrategy; import java.util.ArrayList; import java.util.Random; // ALL OF MY CLASSES import natural.selection.main.animalia.Animal; import natural.selection.main.animalia.OurAnimal; import natural.selection.main.animalia.Predator; import natural.selection.main.animalia.Prey; public class MainApp implements Runnable { private Display display; private int width, height; public String title; private boolean running = false; private Thread thread; private Random random = new Random(); private BufferStrategy bs; private Graphics g; private ArrayList<Animal> allAnimals = new ArrayList<>(); private int[] animalAmt = new int[20]; private ArrayList<Prey> allPreyAnimals = new ArrayList<>(); private int[] preyAnimalAmt = new int[20]; private ArrayList<OurAnimal> allOurAnimals = new ArrayList<>(); private int[] ourAnimalAmt = new int[20]; private ArrayList<Predator> allPredatorAnimals = new ArrayList<>(); private int[] predatorAnimalAmt = new int[20]; public MainApp(String title, int width, int height){ this.width = width; this.height = height; this.title = title; } private void init(){ display = new Display(title, width, height); for(int i=0; i<90; i++) { OurAnimal animal = new OurAnimal(random.nextInt(10), random.nextInt(10), random.nextInt(10), random.nextInt(10), random.nextInt(10), random.nextInt(10)); allOurAnimals.add(animal); allAnimals.add(animal); } for(int i=0; i<(80+random.nextInt(10)); i++) { Prey animal = new Prey(random.nextInt(10), random.nextInt(10), random.nextInt(10), random.nextInt(10), random.nextInt(10), random.nextInt(10)); allPreyAnimals.add(animal); allAnimals.add(animal); } for(int i=0; i<(50+random.nextInt(10)); i++) { Predator animal = new Predator(random.nextInt(10), random.nextInt(10), random.nextInt(10), random.nextInt(10), random.nextInt(10), random.nextInt(10)); allPredatorAnimals.add(animal); allAnimals.add(animal); } animalAmt[0] = allAnimals.size(); preyAnimalAmt[0] = allPreyAnimals.size(); ourAnimalAmt[0] = allOurAnimals.size(); predatorAnimalAmt[0] = allPredatorAnimals.size(); } private void tick() { } // Amount of weeks to simulate private int weeksToSim = 20; private void render(){ bs = display.getCanvas().getBufferStrategy(); if(bs == null){ display.getCanvas().createBufferStrategy(3); return; } g = bs.getDrawGraphics(); //Clear Screen g.clearRect(0, 0, width, height); //Draw Here! g.setColor(Color.black); // LINE GRAPH OF SPECIES COUNT // Draw the graph for(int x=0; x<weeksToSim; x++) { g.drawLine(x*93+125, height-100, x*93+125, 0); g.drawString("Week " + (x+1), x*93+125-20, height-80); } for(int y=0; y<10; y++) { g.drawLine((125), (y+1)*93, width, (y+1)*93); g.drawString(Math.abs(y*10-90) + " alive", (80), (y+1)*93); } // Draw the line for(int i=0; i<(animalAmt.length-1); i++) { // predator animal line g.setColor(Color.red); g.drawLine(i*93+125, (Math.abs((predatorAnimalAmt[i]*10)-90)), (i+1)*93+125, (Math.abs((predatorAnimalAmt[i+1]*10)-90))); g.drawLine(100, height-45, 120, height-45); g.drawString("PREDATOR", 30, height-40); // prey animal line g.setColor(Color.green); g.drawLine(i*93+125, (Math.abs((preyAnimalAmt[i]*10)-90)), (i+1)*93+125, (Math.abs((preyAnimalAmt[i+1]*10)-90))); g.drawLine(100, height-60, 120, height-60); g.drawString("PREY", 30, height-55); // our animal's line g.setColor(Color.blue); g.drawLine(i*93+125, (Math.abs((ourAnimalAmt[i]*10)-90)), (i+1)*93+125, (Math.abs((ourAnimalAmt[i+1]*10)-90))); g.drawLine(100, height-75, 120, height-75); g.drawString("OUR ANIMAL", 30, height-70); } //End Drawing! bs.show(); g.dispose(); } public void run(){ init(); int fps = 60; double timePerTick = 1000000000 / fps; double delta = 0; long now; long lastTime = System.nanoTime(); long timer = 0; int ticks = 0; Toolkit.getDefaultToolkit().sync(); while(running){ now = System.nanoTime(); delta += (now - lastTime) / timePerTick; timer += now - lastTime; lastTime = now; if(delta >= 1){ tick(); render(); ticks++; delta--; } if(timer >= 1000000000){ System.out.println("Ticks and Frames: " + ticks); ticks = 0; timer = 0; } } stop(); } public int getWidth(){ return width; } public int getHeight(){ return height; } public synchronized void start(){ if(running) return; running = true; thread = new Thread(this); thread.start(); } public synchronized void stop(){ if(!running) return; running = false; try { thread.join(); } catch (InterruptedException e) { e.printStackTrace(); } } }
我将如何仅用一个公式使所有要点正确,而又不完全破坏我的所有代码?
这是我的代码所得到的结果:
我要问的基本问题是:我如何制作一个图,其中y = 0处的线实际上位于y = *某个数*,并且随着我的Y的增加,Java的y减少?我是...
让我们将此功能提取为两个函数,这些函数针对给定的x
和y
计算正确的week
和count
位置。这样,您可以通过仅更改一个函数而不是在代码中分散多行来更改图表布局: