MS COCO:指标问题 MAP,MAR: 以项目:mmdetection(v1.0rc0)为例

mac2025-02-27 6

COCO数据集：

{ "info": info, "licenses": [license], "images": [image], "annotations": [annotation], "categories": [category] }

数据集信息

许可

类别：大类-小类(80)

images：40504

annotations：291875

coco.py 中定义了以下 API: COCO 类 - 用于加载 COCO 标注文件并准备所需的数据结构. decodeMask - Decode binary mask M encoded via run-length encoding. encodeMask - Encode binary mask M using run-length encoding. getAnnIds - 获取满足条件的标注信息的 ids. getCatIds - 获取满足条件的类别的 ids. getImgIds - 获取满足条件的图片的 ids. loadAnns - 加载指定 ids 对应的标注信息. loadCats - 加载指定 ids 对应的类别. loadImgs - 加载指定 ids 对应的图片. annToMask - 将 segmentation 标注信息转换为二值 mask. showAnns - 显示指定 ids 的标注信息到对应的图片上. loadRes - 加载算法的结果并创建可用于访问数据的 API . download - 从 mscoco.org 服务器下载 COCO 图片数据集.

imgToAnns: 将图片 id 和标注信息关联;catToImgs: 将类别 id 和图片 id 关联.

这样做的好处就是: 对于给定的图片 id, 就可以快速找到其所有的标注信息; 对于给定的类别 id, 就可以快速找到属于该类的所有图片.

def createIndex(self): """ 创建索引, 属于构造函数的一部分. """ # create index print('creating index...') anns, cats, imgs = {}, {}, {} # imgToAnns: 将图片 id 和标注信息关联; 给定图片 id, 就可以找到其所有的标注信息 # catToImgs: 将类别 id 和图片 id 关联; 给定类别 id, 就可以找到属于这类的所有图片 imgToAnns,catToImgs = defaultdict(list),defaultdict(list) if 'annotations' in self.dataset: for ann in self.dataset['annotations']: imgToAnns[ann['image_id']].append(ann) anns[ann['id']] = ann if 'images' in self.dataset: for img in self.dataset['images']: imgs[img['id']] = img if 'categories' in self.dataset: for cat in self.dataset['categories']: cats[cat['id']] = cat if 'annotations' in self.dataset and 'categories' in self.dataset: for ann in self.dataset['annotations']: catToImgs[ann['category_id']].append(ann['image_id']) print('index created!!!') # 给几个类成员赋值 self.anns = anns self.imgs = imgs self.cats = cats self.imgToAnns = imgToAnns self.catToImgs = catToImgs

关于指标问题:

概要：

https://m.oldpan.me/archives/understand-coco-metric

Pixel Accuracy像素准确度

表示检测物体的准确度，重点判断标准为是否检测到了物体

Pixel Precision像素精度

代表检测到所有的物体中覆盖的精确度，重点判断mask是否精确地覆盖到了该物体。

像素精度是在已经检测到的基础上(不论检测是否失误)进行评测的标准

IOU概念：

https://oldpan.me/archives/iu-iou-intersection-over-union-python

语义分割

https://blog.csdn.net/u014593748/article/details/71698246?fps=1&locationNum=5

检测-分类

https://oldpan.me/archives/understand-coco-metric

https://blog.csdn.net/Gentleman_Qin/article/details/84519388

https://zhuanlan.zhihu.com/p/33273532

https://blog.csdn.net/syoung9029/article/details/56276567#t0

这里有个需求，这边想建立一个统一的标准，去判断mask-rcnn和deeplabv3的算法性能好坏，我是这样设想的，首先是resize应该是同一大小，数据集类别数目统一，数据增强统一，然后mask-rcnn的掩膜作为分割，与GroundTruth对比，计算mIOU，这里阈值取定为0.5，这样就可以统一，和deeplab计算的mIOU去对比，把mask-rcnn当成一个语义分割问题对待，个人想法，不知道是否对

COCO数据集太长了，val2014高达40504张，我们只做一些小测试，修改代码，只载入一部分：

mmdetection/mmdet/datasets/coco.py

def load_annotations(self, ann_file): self.coco = COCO(ann_file) self.cat_ids = self.coco.getCatIds() self.cat2label = { cat_id: i + 1 for i, cat_id in enumerate(self.cat_ids) } self.img_ids = self.coco.getImgIds() img_infos = [] # for i in self.img_ids: # info = self.coco.loadImgs([i])[0] # info['filename'] = info['file_name'] # img_infos.append(info) for key, i in enumerate(self.img_ids): if key == 10: break info = self.coco.loadImgs([i])[0] info['filename'] = info['file_name'] img_infos.append(info) return img_infos

通过枚举函数控制多少个，我们这里只需要10个

接着我们看下图片经过网络后的输出：

mmdetection/tools/test.py

def single_gpu_test(model, data_loader, show=False, ids=None, groundtruth=None, scale_test=None): model.eval() results = [] dataset = data_loader.dataset prog_bar = mmcv.ProgressBar(len(dataset)) for i, data in enumerate(data_loader): with torch.no_grad(): result = model(return_loss=False, rescale=not show, **data)

mmdetection/mmdet/models/detectors/base.py

def show_result(self, data, result, img_norm_cfg, dataset=None, score_thr=0.3, ids=None, groundtruth=None, scale_test=None, ori_hW=None): if isinstance(result, tuple): bbox_result, segm_result = result else: bbox_result, segm_result = result, None

两个都是0-79，共80类

以bbox为例，每一类有很多框或分割：

可视化展示一下第一类的(10,5)

一共五个属性

在mmcv/visualization/image.py中可以知道，是左上x,y 右下x,y 置信度

for bbox, label in zip(bboxes, labels): bbox_int = bbox.astype(np.int32) left_top = (bbox_int[0], bbox_int[1]) right_bottom = (bbox_int[2], bbox_int[3]) cv2.rectangle( img, left_top, right_bottom, bbox_color, thickness=thickness) label_text = class_names[ label] if class_names is not None else 'cls {}'.format(label) if len(bbox) > 4: label_text += '|{:.02f}'.format(bbox[-1]) cv2.putText(img, label_text, (bbox_int[0], bbox_int[1] - 2), cv2.FONT_HERSHEY_COMPLEX, font_scale, text_color)

直接合并成一个array： bboxes = np.vstack(bbox_result)

合并后的array有其对应的lable：

labels = [ np.full(bbox.shape[0], i, dtype=np.int32) for i, bbox in enumerate(bbox_result) ] labels = np.concatenate(labels)

置信度阈值为

score_thr=0.3

大于阈值的显示：

if score_thr > 0: assert bboxes.shape[1] == 5 scores = bboxes[:, -1] inds = scores > score_thr bboxes = bboxes[inds, :] labels = labels[inds] bbox_color = color_val(bbox_color) text_color = color_val(text_color) for bbox, label in zip(bboxes, labels): bbox_int = bbox.astype(np.int32) left_top = (bbox_int[0], bbox_int[1]) right_bottom = (bbox_int[2], bbox_int[3]) cv2.rectangle( img, left_top, right_bottom, bbox_color, thickness=thickness) label_text = class_names[ label] if class_names is not None else 'cls {}'.format(label) if len(bbox) > 4: label_text += '|{:.02f}'.format(bbox[-1]) cv2.putText(img, label_text, (bbox_int[0], bbox_int[1] - 2), cv2.FONT_HERSHEY_COMPLEX, font_scale, text_color)

以segm_result为例：

if segm_result is not None: segms = mmcv.concat_list(segm_result) inds = np.where(bboxes[:, -1] > score_thr)[0] for i in inds: color_mask = np.random.randint( 0, 256, (1, 3), dtype=np.uint8) mask = maskUtils.decode(segms[i]).astype(np.bool) img_show[mask] = img_show[mask] * 0.5 + color_mask * 0.5 segms：

通过阈值0.3判断哪些mask和bbox会被显示：

遍历符合阈值的mask，随机生成颜色(因为是实例分割)：

color_mask = np.random.randint(0, 256, (1, 3), dtype=np.uint8)

这里返回的seg是被编码后的，需要

pycocotools

解码：

mask = maskUtils.decode(segms[i]).astype(np.bool)

返回mask位置

让我们回到mmdetection/tools/test.py

def single_gpu_test(model, data_loader, show=False, ids=None, groundtruth=None, scale_test=None):

返回我们需要的数据结果集合

results

一共10张图像的结果：

第一张：COCO_val2014_000000391895.jpg

configs/mask_rcnn_r50_fpn_1x.py checkpoints/mask_rcnn_r50_fpn_1x_20181010-069fa190.pth --eval bbox segm --out temp_pred/maskrcnn_out.pkl

进行评估：检测和分割

evaluate bbox and segm

先保存到json

def results2json(dataset, results, out_file): result_files = dict() if isinstance(results[0], list): json_results = det2json(dataset, results) result_files['bbox'] = '{}.{}.json'.format(out_file, 'bbox') result_files['proposal'] = '{}.{}.json'.format(out_file, 'bbox') mmcv.dump(json_results, result_files['bbox']) elif isinstance(results[0], tuple): json_results = segm2json(dataset, results) result_files['bbox'] = '{}.{}.json'.format(out_file, 'bbox') result_files['proposal'] = '{}.{}.json'.format(out_file, 'bbox') result_files['segm'] = '{}.{}.json'.format(out_file, 'segm') mmcv.dump(json_results[0], result_files['bbox']) mmcv.dump(json_results[1], result_files['segm']) elif isinstance(results[0], np.ndarray): json_results = proposal2json(dataset, results) result_files['proposal'] = '{}.{}.json'.format(out_file, 'proposal') mmcv.dump(json_results, result_files['proposal']) else: raise TypeError('invalid type of results') return result_files

进入

coco_eval(result_files, eval_types, dataset.coco) def coco_eval(result_files, result_types, coco, max_dets=(100, 300, 1000)): for res_type in result_types: assert res_type in [ 'proposal', 'proposal_fast', 'bbox', 'segm', 'keypoints' ] if mmcv.is_str(coco): coco = COCO(coco) assert isinstance(coco, COCO) if result_types == ['proposal_fast']: ar = fast_eval_recall(result_files, coco, np.array(max_dets)) for i, num in enumerate(max_dets): print('AR@{}\t= {:.4f}'.format(num, ar[i])) return for res_type in result_types: result_file = result_files[res_type] assert result_file.endswith('.json') coco_dets = coco.loadRes(result_file) img_ids = coco.getImgIds() iou_type = 'bbox' if res_type == 'proposal' else res_type cocoEval = COCOeval(coco, coco_dets, iou_type) cocoEval.params.imgIds = img_ids if res_type == 'proposal': cocoEval.params.useCats = 0 cocoEval.params.maxDets = list(max_dets) cocoEval.evaluate() cocoEval.accumulate() cocoEval.summarize()

coco_det:

一共10个图像：

708条记录-bbox+seg：

遍历

for res_type in result_types:

读取

coco_dets = coco.loadRes(result_file)

进入

from pycocotools.cocoeval import COCOeval cocoEval = COCOeval(coco, coco_dets, iou_type)

接着进入：

cocoEval.evaluate() cocoEval.accumulate() cocoEval.summarize()

这里没看懂，先运行一次结果，好低。。。，发现，原来他必须是按照COCO(json)载入在判断，果断用自己的数据集，放弃这种选取10张图片的方式

先接着走流程吧，哎心累

首先是cocoEval.evaluate()

def evaluate(self): ''' Run per image evaluation on given images and store results (a list of dict) in self.evalImgs :return: None ''' tic = time.time() print('Running per image evaluation...') p = self.params # add backward compatibility if useSegm is specified in params if not p.useSegm is None: p.iouType = 'segm' if p.useSegm == 1 else 'bbox' print('useSegm (deprecated) is not None. Running {} evaluation'.format(p.iouType)) print('Evaluate annotation type *{}*'.format(p.iouType)) p.imgIds = list(np.unique(p.imgIds)) if p.useCats: p.catIds = list(np.unique(p.catIds)) p.maxDets = sorted(p.maxDets) self.params=p self._prepare() # loop through images, area range, max detection number catIds = p.catIds if p.useCats else [-1] if p.iouType == 'segm' or p.iouType == 'bbox': computeIoU = self.computeIoU elif p.iouType == 'keypoints': computeIoU = self.computeOks self.ious = {(imgId, catId): computeIoU(imgId, catId) \ for imgId in p.imgIds for catId in catIds} evaluateImg = self.evaluateImg maxDet = p.maxDets[-1] self.evalImgs = [evaluateImg(imgId, catId, areaRng, maxDet) for catId in catIds for areaRng in p.areaRng for imgId in p.imgIds ] self._paramsEval = copy.deepcopy(self.params) toc = time.time() print('DONE (t={:0.2f}s).'.format(toc-tic)) 拿到gt和pred self._prepare() if p.useCats: gts=self.cocoGt.loadAnns(self.cocoGt.getAnnIds(imgIds=p.imgIds, catIds=p.catIds)) dts=self.cocoDt.loadAnns(self.cocoDt.getAnnIds(imgIds=p.imgIds, catIds=p.catIds)) else: gts=self.cocoGt.loadAnns(self.cocoGt.getAnnIds(imgIds=p.imgIds)) dts=self.cocoDt.loadAnns(self.cocoDt.getAnnIds(imgIds=p.imgIds))

参考：

(一) COCO Python API - 使用篇

https://blog.csdn.net/gzj2013/article/details/82385164 (二) COCO Python API - 源码分析篇

https://blog.csdn.net/gzj2013/article/details/82421166

(三) COCO Python API - 数据集类数量分布

https://blog.csdn.net/gzj2013/article/details/82425954

COCO目标检测测评指标

https://www.jianshu.com/p/d7a06a720a2b

计算IOU：

self.ious = {(imgId, catId): computeIoU(imgId, catId) \ for imgId in p.imgIds for catId in catIds}

这里看不到IOU计算的代码，是c语言写的：

https://files.pythonhosted.org/packages/96/84/9a07b1095fd8555ba3f3d519517c8743c2554a245f9476e5e39869f948d2/pycocotools-2.0.0.tar.gz

_mask.pyx

# iou computation. support function overload (RLEs-RLEs and bbox-bbox). def iou( dt, gt, pyiscrowd ): def _preproc(objs): if len(objs) == 0: return objs if type(objs) == np.ndarray: if len(objs.shape) == 1: objs = objs.reshape((objs[0], 1)) # check if it's Nx4 bbox if not len(objs.shape) == 2 or not objs.shape[1] == 4: raise Exception('numpy ndarray input is only for *bounding boxes* and should have Nx4 dimension') objs = objs.astype(np.double) elif type(objs) == list: # check if list is in box format and convert it to np.ndarray isbox = np.all(np.array([(len(obj)==4) and ((type(obj)==list) or (type(obj)==np.ndarray)) for obj in objs])) isrle = np.all(np.array([type(obj) == dict for obj in objs])) if isbox: objs = np.array(objs, dtype=np.double) if len(objs.shape) == 1: objs = objs.reshape((1,objs.shape[0])) elif isrle: objs = _frString(objs) else: raise Exception('list input can be bounding box (Nx4) or RLEs ([RLE])') else: raise Exception('unrecognized type. The following type: RLEs (rle), np.ndarray (box), and list (box) are supported.') return objs def _rleIou(RLEs dt, RLEs gt, np.ndarray[np.uint8_t, ndim=1] iscrowd, siz m, siz n, np.ndarray[np.double_t, ndim=1] _iou): rleIou( <RLE*> dt._R, <RLE*> gt._R, m, n, <byte*> iscrowd.data, <double*> _iou.data ) def _bbIou(np.ndarray[np.double_t, ndim=2] dt, np.ndarray[np.double_t, ndim=2] gt, np.ndarray[np.uint8_t, ndim=1] iscrowd, siz m, siz n, np.ndarray[np.double_t, ndim=1] _iou): bbIou( <BB> dt.data, <BB> gt.data, m, n, <byte*> iscrowd.data, <double*>_iou.data ) def _len(obj): cdef siz N = 0 if type(obj) == RLEs: N = obj.n elif len(obj)==0: pass elif type(obj) == np.ndarray: N = obj.shape[0] return N # convert iscrowd to numpy array cdef np.ndarray[np.uint8_t, ndim=1] iscrowd = np.array(pyiscrowd, dtype=np.uint8) # simple type checking cdef siz m, n dt = _preproc(dt) gt = _preproc(gt) m = _len(dt) n = _len(gt) if m == 0 or n == 0: return [] if not type(dt) == type(gt): raise Exception('The dt and gt should have the same data type, either RLEs, list or np.ndarray') # define local variables cdef double* _iou = <double*> 0 cdef np.npy_intp shape[1] # check type and assign iou function if type(dt) == RLEs: _iouFun = _rleIou elif type(dt) == np.ndarray: _iouFun = _bbIou else: raise Exception('input data type not allowed.') _iou = <double*> malloc(m*n* sizeof(double)) iou = np.zeros((m*n, ), dtype=np.double) shape[0] = <np.npy_intp> m*n iou = np.PyArray_SimpleNewFromData(1, shape, np.NPY_DOUBLE, _iou) PyArray_ENABLEFLAGS(iou, np.NPY_OWNDATA) _iouFun(dt, gt, iscrowd, m, n, iou) return iou.reshape((m,n), order='F')

maskApi.c

void rleIou( RLE *dt, RLE *gt, siz m, siz n, byte *iscrowd, double *o ) { siz g, d; BB db, gb; int crowd; db=malloc(sizeof(double)*m*4); rleToBbox(dt,db,m); gb=malloc(sizeof(double)*n*4); rleToBbox(gt,gb,n); bbIou(db,gb,m,n,iscrowd,o); free(db); free(gb); for( g=0; g<n; g++ ) for( d=0; d<m; d++ ) if(o[g*m+d]>0) { crowd=iscrowd!=NULL && iscrowd[g]; if(dt[d].h!=gt[g].h || dt[d].w!=gt[g].w) { o[g*m+d]=-1; continue; } siz ka, kb, a, b; uint c, ca, cb, ct, i, u; int va, vb; ca=dt[d].cnts[0]; ka=dt[d].m; va=vb=0; cb=gt[g].cnts[0]; kb=gt[g].m; a=b=1; i=u=0; ct=1; while( ct>0 ) { c=umin(ca,cb); if(va||vb) { u+=c; if(va&&vb) i+=c; } ct=0; ca-=c; if(!ca && a<ka) { ca=dt[d].cnts[a++]; va=!va; } ct+=ca; cb-=c; if(!cb && b<kb) { cb=gt[g].cnts[b++]; vb=!vb; } ct+=cb; } if(i==0) u=1; else if(crowd) rleArea(dt+d,1,&u); o[g*m+d] = (double)i/(double)u; } } void bbIou( BB dt, BB gt, siz m, siz n, byte *iscrowd, double *o ) { double h, w, i, u, ga, da; siz g, d; int crowd; for( g=0; g<n; g++ ) { BB G=gt+g*4; ga=G[2]*G[3]; crowd=iscrowd!=NULL && iscrowd[g]; for( d=0; d<m; d++ ) { BB D=dt+d*4; da=D[2]*D[3]; o[g*m+d]=0; w=fmin(D[2]+D[0],G[2]+G[0])-fmax(D[0],G[0]); if(w<=0) continue; h=fmin(D[3]+D[1],G[3]+G[1])-fmax(D[1],G[1]); if(h<=0) continue; i=w*h; u = crowd ? da : da+ga-i; o[g*m+d]=i/u; } } } void rleToBbox( const RLE *R, BB bb, siz n ) { siz i; for( i=0; i<n; i++ ) { uint h, w, x, y, xs, ys, xe, ye, cc, t; siz j, m; h=(uint)R[i].h; w=(uint)R[i].w; m=R[i].m; m=((siz)(m/2))*2; xs=w; ys=h; xe=ye=0; cc=0; if(m==0) { bb[4*i+0]=bb[4*i+1]=bb[4*i+2]=bb[4*i+3]=0; continue; } for( j=0; j<m; j++ ) { cc+=R[i].cnts[j]; t=cc-j%2; y=t%h; x=(t-y)/h; xs=umin(xs,x); xe=umax(xe,x); ys=umin(ys,y); ye=umax(ye,y); } bb[4*i+0]=xs; bb[4*i+2]=xe-xs+1; bb[4*i+1]=ys; bb[4*i+3]=ye-ys+1; } }

通过代码我们简单的看出，seg的IOU计算，是对整个图像，也就是n*m的矩阵进行对比，当然这里细节还不理解，比如，一个图里有很多类，每一个类有很多框，对应GT也是如此

那么怎么匹配的计算IOU

最新回复(0)