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按矩形坐标对颜色排序

colors sorting java

ProgrammingCuber · 技术社区 · 7 年前

我正在开发一个程序,在这个程序中,我拍摄了一张照片,找到照片中的正方形,然后确定它们的RGB值。我能够成功地获得这些信息,但当我找到RGB值和坐标时,它们按照我找到的轮廓的顺序排列。我的问题是,我怎样才能把颜色分类,让我像看书一样从左上角一行读到底行。每种颜色都有矩形坐标,我想用它来帮我排序。

我有一张图片,上面显示了我想要点的东西。

当我从图像中获得RGB值时,我打印出颜色和坐标。下面是它的样子(注意它们是如何被订购的):

279,271
R: 106 G: 255 B: 183 

188,270
R: 107 G: 255 B: 180 

371,267
R: 102 G: 255 B: 169 

274,178
R: 75 G: 255 B: 99 

185,179
R: 84 G: 255 B: 103 

369,175
R: 89 G: 255 B: 105 

277,88
R: 96 G: 255 B: 103 

184,85
R: 101 G: 255 B: 110 

370,84
R: 124 G: 255 B: 126

我用来获取它们的代码如下:

我只是不知道怎样才能把颜色分类存储。

private void getColors(Mat img2read , Rect roi){
    int rAvg = 0 , bAvg = 0, gAvg = 0;
    int rSum = 0, bSum = 0, gSum = 0;

    img2read = new Mat(img2read, roi); //image size of rect (saved contour size and coordinates)
    img2read.convertTo(img2read, CvType.CV_8UC1); //convert to workable type for getting RGB data

    int totalBytes = (int) (img2read.total() * img2read.channels());
    byte buff[] = new byte[totalBytes];
    int channels = img2read.channels();

    img2read.get(0,0,buff);
    int stride = channels * img2read.width();
    for(int i = 0; i < img2read.height();i++){
        for(int x = 0; x < stride; x+= channels){
            int r = unsignedToBytes(buff[(i * stride) + x]);
            int g = unsignedToBytes(buff[(i * stride)+ x + 1]);
            int b = unsignedToBytes(buff[(i * stride)+ x + 2]);

            rSum += r;
            gSum += g;
            bSum += b;

        }
    }
    float[] hsv = new float[3];

    rAvg = (int) (rSum / img2read.total());
    gAvg = (int) (gSum /  img2read.total());
    bAvg = (int) (bSum / img2read.total());


    Color.RGBtoHSB(rAvg, gAvg, bAvg, hsv);

    hsv[2] = 1; //Set to max value

    int rgb = Color.HSBtoRGB(hsv[0], hsv[1], hsv[2]);

    Color brightenedColor = new Color(rgb);
    System.out.println();
    System.out.println(roi.x + "," + roi.y);
    System.out.printf("R: %s G: %s B: %s \n", brightenedColor.getRed(), brightenedColor.getGreen(), brightenedColor.getBlue());     
    Color colorToAdd = new Color(brightenedColor.getRed(), brightenedColor.getGreen(), brightenedColor.getBlue());
    Point coords = new Point(roi.x, roi.y);
    //colorCoords.put(coords, colorToAdd);
    /*counter++;
    if(counter == 9){
        sortColors(colorCoords);
    }*/
}

int counter = 0;

Map <Point, Color> colorCoords = new TreeMap<>();
Color[] colorArray = new Color[54];
int currentIndex = 0;

//Not really sure what I'm doing here but I don't think it is right.
private void sortColors(Map<Point, Color> colorCoords){

    for(int i = 0; i < colorCoords.size();i++){
    //  colorCoords.get(key)
    }

}

2 回复 | 直到 7 年前

Usagi Miyamoto 7 年前

将这些值放入对象中,并将其存储到 List . 要排序,可以添加静态 Comparator 字段,或使其实现 Comparable 类似这样:

public class Block implement Comparable<Block> {
   public int x, y;
   public Color color;
   @Override
   public int compareTo(Block other) {
      if (null==other) throw new NullPointerException();
      // implement the logic, like:
      // this will compare X, unless difference in Y is more than EG 10
      return Math.abs(y - other.y) > 10 ? y - other.y : x - other.x;
   }
   public Color getColor() { return color; }
}

public List<Block> blocks = new ArrayList<>();

要排序,可以使用

Collections.sort(blocks);

编辑了上述比较,以满足您的需要。。。

Java8 的得到一个排序的数组应该是可行的 Color

 Color[] colors = blocks.stream().sorted().map(Block::getColor).toArray();

为了让它发挥作用 Block.getColor()

Jason Trader 7 年前