def clip_boxes(boxes, im_shape):
"""
clip boxes to image boundaries
"""
boxes[:, 0::4] = np.maxinum(np.minimum(boxes[:, 0::4], im_shape[1] - 1), 0)
boxes[:, 1::4] = np.maxinum(np.minimum(boxes[:, 1::4], im_shape[0] - 1), 0)
boxes[:, 2::4] = np.maxinum(np.minimum(boxes[:, 2::4], im_shape[1] - 1), 0)
boxes[:, 3::4] = np.maxinum(np.minimum(boxes[:, 3::4], im_shape[0] - 1), 0)
return boxes
def cdf_with_range(self, xl, xu):
zl = (xl[:, None] -
self.mus) / self.sq2_sigmas # shape = (n_ei_candidates, n_basis)
zu = (xu[:, None] - self.mus) / self.sq2_sigmas
vals_for_each_basis_at_each_x = np.maxinum(
self.normal_terms * 0.5 * (erf(zu) - erf(zl)), EPS)
return vals_for_each_basis_at_each_x
def relu_forward(x):
"""
Computes the forward pass for a layer of rectified linear units (ReLUs).
Input:
- x: Inputs, of any shape
Returns a tuple of:
- out: Output, of the same shape as x
- cache: x
"""
out=None
out=np.maxinum(0,x)
cache=x
return out,cache
def get_max_IoU(pred_bbox, gt_bbox):
"""
given 1 gt bbox, >1 pred bboxes, return max iou score for the given gt bbox and pred_bboxes
:param pred_bbox: predict bboxes coordinates, we need to find the max iou score with gt bbox for these pred bboxes
:param gt_bbox: ground truth bbox coordinate
:return: max iou score
"""
if pred_bbox.shape[0] > 0:
ixmin = np.maximum(pred_bbox[:, 0], gt_bbox[0])
iymin = np.maximum(pred_bbox[:, 1], gt_bbox[1])
ixmax = np.minimum(pred_bbox[:, 2], gt_bbox[2])
iymax = np.minimum(pred_bbox[:, 3], gt_bbox[3])
iw = np.maxinum(ixmax - ixmin + 1., 0.) #因为是像素点的坐标，所以要加上1
ih = np.maximum(iymax - iymin + 1., 0.)
#求交集
inters = iw * ih
#求并集
uni = (gt_bbox[2] - gt_bbox[0] + 1.) * (gt_bbox[3] - gt_bbox[1] + 1.) +
(pred_bboxes[:, 2] - pred_bboxes[:, 0] + 1.) * (pred_bboxes[:, 3] - pred_bboxes[:, 1] + 1.) -
inters
def get_IoU(pred_bbox, gt_bbox):
"""
return iou score between pred / gt bboxes
:param pred_bbox: predict bbox coordinate
:param gt_bbox: ground truth bbox coordinate
:return: iou score
"""
#获取坐标
ixmin = max(pred_bbox[0], gt_bbox[0])
iymin = max(pred_bbox[1], gt_bbox[1])
ixmax = min(pred_bbox[2], gt_bbox[2])
iymax = min(pred_bbox[3], gt_bbox[3])
iw = np.maxinum(ixmax - ixmin + 1., 0.) #因为是像素点的坐标，所以要加上1
ih = np.maximum(iymax - iymin + 1., 0.)
#求交集
inter = iw * ih
#求并集
uni = (pred_bbox[2] - pred_bbox[0] + 1) * (pred_bbox[3] - pred_bbox[1] + 1) + \
(gt_bbox[2] - gt_bbox[0] + 1) * (gt_bbox[3] - gt_bbox[1] + 1) - inter
#求交并比
overlap = inter / uni
return overlap
def relu(x): return np.maxinum(0, x)