模拟量差分和单端(iou计算方法)

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一．计算IOU

def intersect(box_a, box_b):
    max_xy = np.minimum(box_a[:, 2:], box_b[2:])
    min_xy = np.maximum(box_a[:, :2], box_b[:2])
    inter = np.clip((max_xy - min_xy), a_min=0, a_max=np.inf)
    return inter[:, 0] * inter[:, 1]


def jaccard_numpy(box_a, box_b):
    """Compute the jaccard overlap of two sets of boxes.  The jaccard overlap
    is simply the intersection over union of two boxes.
    E.g.:
    Args:
        box_a: Multiple bounding boxes, Shape: [num_boxes,4]
        box_b: Single bounding box, Shape: [4]
    Return:
        jaccard overlap: Shape: [box_a.shape[0], box_a.shape[1]]
    """
    inter = intersect(box_a, box_b)
    area_a = ((box_a[:, 2]-box_a[:, 0]) *
              (box_a[:, 3]-box_a[:, 1]))  # [A,B]
    area_b = ((box_b[2]-box_b[0]) *
              (box_b[3]-box_b[1]))  # [A,B]
    union = area_a + area_b - inter
    return inter / union  # [A,B]

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人工修正后的每一个框与算法输出的所有框去计算IOU,取出IOU大于０.９的算法输出框

def compute_IOU(algrim_bboxs,fix_bboxs):
    # print('algrim_bboxs:', algrim_bboxs)
    # print('fix_bboxs:', fix_bboxs)
    for i, fix_bbox in enumerate(fix_bboxs):
        print('fix_bbox:', fix_bbox)
        fix_x1, fix_y1, fix_x2, fix_y2 = fix_bbox
        x1, y1, x2, y2 = algrim_bboxs[:,0], algrim_bboxs[:,1], algrim_bboxs[:,-2], algrim_bboxs[:,-1]
        area = (y2-y1)*(x2-x1)
        inter = np.maximum((np.minimum(x2,fix_x2) - np.maximum(x1,fix_x1)),0)*np.maximum((np.minimum(y2,fix_y2) - np.maximum(y1, fix_y1)),0)
        union = area+(fix_y2-fix_y1)*(fix_x2-fix_x1)-inter
        IOU = inter/union
        # print('IOU:', IOU)
        index = np.where(IOU > 0.9)[0]
        print('algrim_bboxs[index]:', algrim_bboxs[index])

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注释：下面的代码假设了第0行是IOU最大的,那么其余做NMS计算

boxes=np.array([[1,2,3,4],
              [5,6,7,8],
              [9,10,11,12]])
y1 = boxes[:, 0]
x1 = boxes[:, 1]
y2 = boxes[:, 2]
x2 = boxes[:, 3]
area = (y2 - y1) * (x2 - x1)
print('area=',area)
box_area=area[0]
print('box_area=',box_area)
boxes_area=area[1:]
print('boxes_area=',boxes_area)
box=boxes[0]
y1 = np.maximum(box[0], boxes[1:, 0])
y2 = np.minimum(box[2], boxes[1:, 2])
x1 = np.maximum(box[1], boxes[1:, 1])
x2 = np.minimum(box[3], boxes[1:, 3])
#注意与0判断 若不想交返回0值，即intersection为0
intersection = np.maximum(x2 - x1, 0) * np.maximum(y2 - y1, 0)
print('intersection=',intersection)
union = box_area + boxes_area - intersection[:]
print('union=',union)
iou = intersection / union
print('iou=',iou)

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二，一般加上score，这样用

scores = np.array([0.5, 0.7, 0.3, 0.2])
ixs=scores.argsort()[::-1]
print('ixs=',ixs)
boxes=np.array([[1,2,3,4],
                [1.2, 0.4,2.1, 0.8],
                [5,6,7,8],
                [9,10,11,12]])
y1 = boxes[:, 0]
x1 = boxes[:, 1]
y2 = boxes[:, 2]
x2 = boxes[:, 3]
area = (y2 - y1) * (x2 - x1)
print('area=',area)
box_area=area[ixs[0]]
print('box_area=',box_area)
boxes_area=area[ixs[1:]]
print('boxes_area=',boxes_area)
box=boxes[ixs[0]]
boxes=boxes[ixs[1:]]
y1 = np.maximum(box[0], boxes[:, 0])
y2 = np.minimum(box[2], boxes[:, 2])
x1 = np.maximum(box[1], boxes[:, 1])
x2 = np.minimum(box[3], boxes[:, 3])
#注意与0判断 若不想交返回0值，即intersection为0
intersection = np.maximum(x2 - x1, 0) * np.maximum(y2 - y1, 0)
print('intersection=',intersection)
union = box_area + boxes_area - intersection[:]
print('union=',union)
iou = intersection / union
print('iou=',iou)

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三，求其NMS后的框的小trick

scores=np.array([0.8,0.4,0.7,0.2,0.5])
ixs = scores.argsort()[::-1]
print('ixs=',ixs)
#iou要比score少一个数值，因为iou的计算是由score的第一个值确定的，故下面的np.where要加1
iou=np.array([0.3,0.6,0.1,0.4])
threshold=0.3
remove_ixs = np.where(iou > threshold)[0]+1
print('remove_ixs=',remove_ixs)
# Remove indices of the picked and overlapped boxes.
ixs = np.delete(ixs, remove_ixs)
print('ixs=',ixs)
ixs = np.delete(ixs, 0)
print('ixs=',ixs)

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四，加入while，就把不要的框删掉

scores = np.array([0.8, 0.4, 0.7, 0.2, 0.5])
ixs = scores.argsort()[::-1]
while len(ixs) > 0:
    print('================================')
    print('ixs=', ixs)
    # iou要比score少一个数值，因为iou的计算是由score的第一个值确定的，故下面的np.where要加1
    iou = np.array([0.3, 0.6, 0.1, 0.4])
    threshold = 0.3
    remove_ixs = np.where(iou > threshold)[0] + 1
    print('remove_ixs=', remove_ixs)
    # Remove indices of the picked and overlapped boxes.
    ixs = np.delete(ixs, remove_ixs)
    print('ixs=', ixs)
    ixs = np.delete(ixs, 0)
    print('ixs=', ixs)

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五，faster rcnn NMS

def py_cpu_nms(dets, thresh):
    """Pure Python NMS baseline."""
    x1 = dets[:, 0]
    y1 = dets[:, 1]
    x2 = dets[:, 2]
    y2 = dets[:, 3]
    scores = dets[:, 4]

    areas = (x2 - x1 + 1) * (y2 - y1 + 1)
    order = scores.argsort()[::-1]
    print('order',order)
    keep = []
    while order.size > 0:
        i = order[0]
        keep.append(i)
        xx1 = np.maximum(x1[i], x1[order[1:]])
        yy1 = np.maximum(y1[i], y1[order[1:]])
        xx2 = np.minimum(x2[i], x2[order[1:]])
        yy2 = np.minimum(y2[i], y2[order[1:]])

        w = np.maximum(0.0, xx2 - xx1 + 1)
        h = np.maximum(0.0, yy2 - yy1 + 1)
        inter = w * h
        ovr = inter / (areas[i] + areas[order[1:]] - inter)

        print('ovr=',ovr)
        print('np.where(ovr <= thresh)',np.where(ovr <= thresh))
        inds = np.where(ovr <= thresh)[0]
        print('inds=',inds)
        print('inds + 1',inds + 1)
        print('order[inds + 1]',order[inds + 1])
        order = order[inds + 1]
        print('order=',order)

    return keep
def nms():
    # ###self nms
    result = [np.array([[7.9301813e+02, 6.3052429e+02, 8.9795898e+02,             
                 7.2513965e+02,9.9307442e-01],
                     [8.0682990e+02, 6.4606281e+02, 8.7198071e+02, 7.1328979e+02,
                      9.5732883e-02]], dtype=np.float32)]
    ##faster rcnn  nms
    keep=py_cpu_nms(result[0],0.7)
    print('keep=',keep)
    for i in keep:
        print('i=',i)
        print(result[0][i])

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此处大于0.7IOU的框就抑制, 小于0.7的就留下.

六．多边形利用polygon计算IOU

def compute_IOU_():
    import numpy as np
    import shapely
    from shapely.geometry import Polygon, MultiPoint  # 多边形

    path = './134.jpg'
    img = cv2.imread(path)
    print('img.shape:', img.shape)
    line1 = [728, 252, 908, 215, 934, 312, 752, 355]  # 四边形四个点坐标的一维数组表示，[x,y,x,y....]
    line2 = [741, 262, 907, 228, 923, 308, 758, 342]

    # #debug to show
    # line1 = np.array(line1).reshape(4, 2)
    # line2 = np.array(line2).reshape(4, 2)
    # cv2.polylines(img, [np.array(line1).reshape(-1, 1, 2)], True, (0, 255, 0), thickness=2)
    # cv2.polylines(img, [np.array(line2).reshape(-1, 1, 2)], True, (0, 0, 255), thickness=2)
    # cv2.imwrite('134_with_rect.jpg',img)

    line1_box = np.array(line1).reshape(4, 2)  # 四边形二维坐标表示
    # 凸多边形
    # poly1 = Polygon(line1_box).convex_hull
    #凸多边形与凹多边形
    poly1 = Polygon(line1_box)#.convex_hull  # python四边形对象，会自动计算四个点，最后四个点顺序为：左上 左下  右下 右上 左上
    print(Polygon(line1_box).convex_hull)
    line2_box = np.array(line2).reshape(4, 2)
    # 凸多边形
    # poly2 = Polygon(line2_box).convex_hull
    # 凸多边形与凹多边形
    poly2 = Polygon(line2_box)
    print(Polygon(line2_box).convex_hull)

    union_poly = np.concatenate((line1_box, line2_box))  # 合并两个box坐标，变为8*2
    # print(union_poly)
    print(MultiPoint(union_poly).convex_hull)  # 包含两四边形最小的多边形点
    if not poly1.intersects(poly2):  # 如果两四边形不相交
        iou = 0
    else:
        try:
            inter_area = poly1.intersection(poly2).area  # 相交面积
            print(inter_area)
            union_area = MultiPoint(union_poly).convex_hull.area
            print(union_area)
            if union_area == 0:
                iou = 0
            # iou = float(inter_area) / (union_area-inter_area)  #错了
            iou = float(inter_area) / union_area
        except shapely.geos.TopologicalError:
            print('shapely.geos.TopologicalError occured, iou set to 0')
            iou = 0
    print('line1_box:', line1_box)
    print('iou:', iou)

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