二值mask转polygon/RLE (coco segment格式）

coco数据集annotation的segmentation并不是二值mask，而是polygon格式，
看一个annotation.

{
	"segmentation": [[510.66,423.01,511.72,420.03,510.45......]], #两两组成(x,y)坐标，polygon格式
	"area": 702.1057499999998, #面积
	"iscrowd": 0,  #是不是一群物体,为0是seg是polygon格式，否则是RLE格式
	"image_id": 289343,  #对应的image id
	"bbox": [473.07,395.93,38.65,28.67], #(x,y,w,h)
	"category_id": 18,  #分类label
	"id": 1768  #当前annotation的id,每一个图像有不止一个对象，所以要对每一个对象编号（每个对象的ID是唯一的）
},

segmentation其实是一个二值mask的轮廓点，
如果想把二值mask转成这种格式，需要提取轮廓。

1.Polygon格式

网上找来一张二值mask的图片

提取它的轮廓

mask_img = cv2.imread("mask.png",cv2.IMREAD_GRAYSCALE)
contours, _ = cv2.findContours(mask_img, cv2.RETR_TREE, cv2.CHAIN_APPROX_NONE)
polygons = []
for object in contours:
    coords = []

    for point in object:
        coords.append(int(point[0][0]))
        coords.append(int(point[0][1]))

    polygons.append(coords)
    print(polygons)
    #[[131, 48, 130, 49, 129, 50, 128, 51, ...]]

这个polygon能不能用呢？是不是和coco的annotation格式一样？
下面来验证一下。
引入python自带的coco库，要用到里面的annToMask函数，把polygon转为mask,
然后imshow出来看看是不是和原图一样。

先看一下annToMask函数，它需要先把polygon格式转为RLE格式。

    def annToMask(self, ann):
        """
        Convert annotation which can be polygons, uncompressed RLE, or RLE to binary mask.
        :return: binary mask (numpy 2D array)
        """
        rle = self.annToRLE(ann)
        m = maskUtils.decode(rle)
        return m
        
    def annToRLE(self, ann):
        """
        Convert annotation which can be polygons, uncompressed RLE to RLE.
        :return: binary mask (numpy 2D array)
        """
        t = self.imgs[ann['image_id']]
        h, w = t['height'], t['width']
        segm = ann['segmentation']
        if type(segm) == list:
            # polygon -- a single object might consist of multiple parts
            # we merge all parts into one mask rle code
            rles = maskUtils.frPyObjects(segm, h, w)
            rle = maskUtils.merge(rles)
        elif type(segm['counts']) == list:
            # uncompressed RLE
            rle = maskUtils.frPyObjects(segm, h, w)
        else:
            # rle
            rle = ann['segmentation']
        return rle

转成RLE的过程中需要用到annotation里的image_id, segmentation.
而现在我们只有一个segmentation, 并没有image_id,
而image_id只是用来得到图片的w, h，
所以引入coco的instance_val2017.json文件，随便找一个image_id用一下。

from pycocotools.coco import COCO
coco_api = COCO('coco/annotations/instances_val2017.json')
ann = dict()
ann['image_id'] = 37777
ann['segmentation'] = polygons
mask = coco_api.annToMask(ann)
mask = np.clip(mask*255,0,255)

说明找到的轮廓是可以用的（随便找的image_id导致h，w和原图不一致，不过不影响mask）.
（在曲线的连续性上仔细看会有一些误差）。

还是首选下面的RLE格式。

2.RLE格式

如果用在coco的segment, 首选这种，上面轮廓处会存在误差，且coco函数中polygon格式有诸多限制。

mask转RLE的代码来自segment anything的utils，
这里的mask是tensor形式的二值mask, 值是int型。

    def mask_to_rle(tensor: torch.Tensor) -> List[Dict[str, Any]]:
        """
        Encodes masks to an uncompressed RLE, in the format expected by
        pycoco tools.
        """
        # Put in fortran order and flatten h,w
        h, w, b = tensor.shape  #需要根据tensor的shape修改
        tensor = tensor.permute(2, 1, 0).flatten(1)  #需要根据tensor的shape修改

        # Compute change indices
        diff = tensor[:, 1:] ^ tensor[:, :-1] #要求值是int型
        change_indices = diff.nonzero()

        # Encode run length
        out = []
        for i in range(b):
            cur_idxs = change_indices[change_indices[:, 0] == i, 1]
            cur_idxs = torch.cat(
                [
                    torch.tensor([0], dtype=cur_idxs.dtype, device=cur_idxs.device),
                    cur_idxs + 1,
                    torch.tensor([h * w], dtype=cur_idxs.dtype, device=cur_idxs.device),
                ]
            )
            btw_idxs = cur_idxs[1:] - cur_idxs[:-1]
            counts = [] if tensor[i, 0] == 0 else [0]
            counts.extend(btw_idxs.detach().cpu().tolist())
            out.append({"size": [h, w], "counts": counts})
        return out