来自 https://zhuanlan.zhihu.com/p/47189358
IoU（Intersection over Union）交并比
ground truth bbox、predict bbox交集的面积 / 二者并集的面积；
主要看代码实现
import numpy as np
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
"""
# bbox should be valid, actually we should add more judgements, just ignore here...
# assert ((abs(pred_bbox[2] - pred_bbox[0]) > 0) and
# (abs(pred_bbox[3] - pred_bbox[1]) > 0))
# assert ((abs(gt_bbox[2] - gt_bbox[0]) > 0) and
# (abs(gt_bbox[3] - gt_bbox[1]) > 0))
# -----0---- get coordinates of inters
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.maximum(ixmax - ixmin + 1., 0.)
ih = np.maximum(iymax - iymin + 1., 0.)
# -----1----- intersection
inters = iw * ih
# -----2----- union, uni = S1 + S2 - inters
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.) -
inters)
# -----3----- iou
overlaps = inters / uni
return overlaps
def get_max_IoU(pred_bboxes, gt_bbox):
"""
given 1 gt bbox, >1 pred bboxes, return max iou score for the given gt bbox and pred_bboxes
:param pred_bboxes: 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
"""
# bbox should be valid, actually we should add more judgements, just ignore here...
# assert ((abs(gt_bbox[2] - gt_bbox[0]) > 0) and
# (abs(gt_bbox[3] - gt_bbox[1]) > 0))
if pred_bboxes.shape[0] > 0:
# -----0---- get coordinates of inters, but with multiple predict bboxes
ixmin = np.maximum(pred_bboxes[:, 0], gt_bbox[0])
iymin = np.maximum(pred_bboxes[:, 1], gt_bbox[1])
ixmax = np.minimum(pred_bboxes[:, 2], gt_bbox[2])
iymax = np.minimum(pred_bboxes[:, 3], gt_bbox[3])
iw = np.maximum(ixmax - ixmin + 1., 0.)
ih = np.maximum(iymax - iymin + 1., 0.)
# -----1----- intersection
inters = iw * ih
# -----2----- union, uni = S1 + S2 - inters
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)
# -----3----- iou, get max score and max iou index
overlaps = inters / uni
ovmax = np.max(overlaps) #max iou score
jmax = np.argmax(overlaps) #max iou index
return overlaps, ovmax, jmax
if __name__ == "__main__":
# test1
pred_bbox = np.array([50, 50, 90, 100]) # top-left: <50, 50>, bottom-down: <90, 100>, <x-axis, y-axis>
gt_bbox = np.array([70, 80, 120, 150])
print (get_IoU(pred_bbox, gt_bbox))
# test2
pred_bboxes = np.array([[15, 18, 47, 60],
[50, 50, 90, 100],
[70, 80, 120, 145],
[130, 160, 250, 280],
[25.6, 66.1, 113.3, 147.8]])
gt_bbox = np.array([70, 80, 120, 150])
print (get_max_IoU(pred_bboxes, gt_bbox))
import numpy as np
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
"""
# bbox should be valid, actually we should add more judgements, just ignore here...
# assert ((abs(pred_bbox[2] - pred_bbox[0]) > 0) and
# (abs(pred_bbox[3] - pred_bbox[1]) > 0))
# assert ((abs(gt_bbox[2] - gt_bbox[0]) > 0) and
# (abs(gt_bbox[3] - gt_bbox[1]) > 0))
# -----0---- get coordinates of inters
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.maximum(ixmax - ixmin + 1., 0.)
ih = np.maximum(iymax - iymin + 1., 0.)
# -----1----- intersection
inters = iw * ih
# -----2----- union, uni = S1 + S2 - inters
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.) -
inters)
# -----3----- iou
overlaps = inters / uni
return overlaps
def get_max_IoU(pred_bboxes, 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
"""
# bbox should be valid, actually we should add more judgements, just ignore here...
# assert ((abs(gt_bbox[2] - gt_bbox[0]) > 0) and
# (abs(gt_bbox[3] - gt_bbox[1]) > 0))
if pred_bboxes.shape[0] > 0:
# -----0---- get coordinates of inters, but with multiple predict bboxes
ixmin = np.maximum(pred_bboxes[:, 0], gt_bbox[0])
iymin = np.maximum(pred_bboxes[:, 1], gt_bbox[1])
ixmax = np.minimum(pred_bboxes[:, 2], gt_bbox[2])
iymax = np.minimum(pred_bboxes[:, 3], gt_bbox[3])
iw = np.maximum(ixmax - ixmin + 1., 0.)
ih = np.maximum(iymax - iymin + 1., 0.)
# -----1----- intersection
inters = iw * ih
# -----2----- union, uni = S1 + S2 - inters
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)
# -----3----- iou, get max score and max iou index
overlaps = inters / uni
ovmax = np.max(overlaps)
jmax = np.argmax(overlaps)
return overlaps, ovmax, jmax
if __name__ == "__main__":
# test1
pred_bbox = np.array([50, 50, 90, 100]) # top-left: <50, 50>, bottom-down: <90, 100>, <x-axis, y-axis>
gt_bbox = np.array([70, 80, 120, 150])
print (get_IoU(pred_bbox, gt_bbox))
# test2
pred_bboxes = np.array([[15, 18, 47, 60],
[50, 50, 90, 100],
[70, 80, 120, 145],
[130, 160, 250, 280],
[25.6, 66.1, 113.3, 147.8]])
gt_bbox = np.array([70, 80, 120, 150])
print (get_max_IoU(pred_bboxes, gt_bbox))原文：https://www.cnblogs.com/mengting-123/p/13704900.html