Posted in Python onJanuary 18, 2020
解决问题: 不使用for计算两组、多个矩形两两间的iou
使用numpy广播的方法,在python程序中并不建议使用for语句,python中的for语句耗时较多,如果使用numpy广播的思想将会提速不少。
代码:
def calc_iou(bbox1, bbox2): if not isinstance(bbox1, np.ndarray): bbox1 = np.array(bbox1) if not isinstance(bbox2, np.ndarray): bbox2 = np.array(bbox2) xmin1, ymin1, xmax1, ymax1, = np.split(bbox1, 4, axis=-1) xmin2, ymin2, xmax2, ymax2, = np.split(bbox2, 4, axis=-1) area1 = (xmax1 - xmin1) * (ymax1 - ymin1) area2 = (xmax2 - xmin2) * (ymax2 - ymin2) ymin = np.maximum(ymin1, np.squeeze(ymin2, axis=-1)) xmin = np.maximum(xmin1, np.squeeze(xmin2, axis=-1)) ymax = np.minimum(ymax1, np.squeeze(ymax2, axis=-1)) xmax = np.minimum(xmax1, np.squeeze(xmax2, axis=-1)) h = np.maximum(ymax - ymin, 0) w = np.maximum(xmax - xmin, 0) intersect = h * w union = area1 + np.squeeze(area2, axis=-1) - intersect return intersect / union
程序中输入为多个矩形[xmin, ymin, xmax,ymax]格式的数组或者list,输出为numpy格式,例:输入的shape为(3, 4)、(5,4)则输出为(3, 5)各个位置为boxes间相互的iou值。后面会卡一个iou的阈值,然后就可以将满足条件的索引取出。如:
def delete_bbox(bbox1, bbox2, roi_bbox1, roi_bbox2, class1, class2, idx1, idx2, iou_value): idx = np.where(iou_value > 0.4) left_idx = idx[0] right_idx = idx[1] left = roi_bbox1[left_idx] right = roi_bbox2[right_idx] xmin1, ymin1, xmax1, ymax1, = np.split(left, 4, axis=-1) xmin2, ymin2, xmax2, ymax2, = np.split(right, 4, axis=-1) left_area = (xmax1 - xmin1) * (ymax1 - ymin1) right_area = (xmax2 - xmin2) * (ymax2 - ymin2) left_idx = left_idx[np.squeeze(left_area < right_area, axis=-1)]#小的被删 right_idx = right_idx[np.squeeze(left_area > right_area, axis=-1)] bbox1 = np.delete(bbox1, idx1[left_idx], 0) class1 = np.delete(class1, idx1[left_idx]) bbox2 = np.delete(bbox2, idx2[right_idx], 0) class2 = np.delete(class2, idx2[right_idx]) return bbox1, bbox2, class1, class2
IOU计算原理:
ymin = np.maximum(ymin1, np.squeeze(ymin2, axis=-1)) xmin = np.maximum(xmin1, np.squeeze(xmin2, axis=-1)) ymax = np.minimum(ymax1, np.squeeze(ymax2, axis=-1)) xmax = np.minimum(xmax1, np.squeeze(xmax2, axis=-1)) h = np.maximum(ymax - ymin, 0) w = np.maximum(xmax - xmin, 0) intersect = h * w
计算矩形间min的最大值,max的最小值,如果ymax-ymin值大于0则如左图所示,如果小于0则如右图所示
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python不使用for计算两组、多个矩形两两间的iou方式
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