交并比(Intersection-over-Union,IoU),目的检测中运用的一个概念,我们在停止目的检测算法测试时,重要的目的,是产生的预测框(candidate bound)与标记框(ground truth bound)的交叠率,即它们的交集与并集的比值。最理想状况是完全堆叠,即比值为1。
通常,我们所说的目的检测检测的框是规则的矩形框,计算IOU也十分复杂,普通两种办法:
两个矩形的宽之和减去组合后的矩形的宽就是堆叠矩形的宽,同比堆叠矩形的高。
右下角的最小值减去左上角的最大值就是堆叠矩形的宽,同比高。
上述规则四边形(矩形)IOU计算方式一的 Python完成
def calculate_regular_iou(rec1, rec2):
"""
computing IoU
:param rec1: (y0, x0, y1, x1), which reflects
(top, left, bottom, right)
:param rec2: (y0, x0, y1, x1)
:return: scala value of IoU
"""
S_rec1 = (rec1[2] - rec1[0]) * (rec1[3] - rec1[1])
S_rec2 = (rec2[2] - rec2[0]) * (rec2[3] - rec2[1])
sum_area = S_rec1 + S_rec2
left_line = max(rec1[1], rec2[1])
right_line = min(rec1[3], rec2[3])
top_line = max(rec1[0], rec2[0])
bottom_line = min(rec1[2], rec2[2])
if left_line >= right_line or top_line >= bottom_line:
return 0
else:
intersect = (right_line - left_line) * (bottom_line - top_line)
return (intersect / (sum_area - intersect)) * 1.0
if __name__ == '__main__':
# (top, left, bottom, right)
rect1 = [551, 26, 657, 45]
rect2 = [552, 27, 672, 46]
iou = calculate_regular_iou(rect1, rect2)
上述规则四边形(矩形)IOU计算方式二的 Python 完成
def compute_regular_iou_other(rec1, rec2):
"""
computing IoU
:param rec1: (y0, x0, y1, x1), which reflects
(top, left, bottom, right)
:param rec2: (y0, x0, y1, x1)
:return: scala value of IoU
"""
areas1 = (rec1[3] - rec1[1]) * (rec1[2] - rec1[0])
areas2 = (rec2[3] - rec2[1]) * (rec2[2] - rec2[0])
left = max(rec1[1],rec2[1])
right = min(rec1[3],rec2[3])
top = max(rec1[0], rec2[0])
bottom = min(rec1[2], rec2[2])
w = max(0, right - left)
h = max(0, bottom - top)
return w*h / (areas2 + areas1 - w*h)
if __name__ == '__main__':
# (top, left, bottom, right)
rect1 = [551, 26, 657, 45]
rect2 = [552, 27, 672, 46]
iou = compute_regular_iou_other(rect1, rect2)
但是,关于不规则四边形就不能经过上述这两种方式来计算,这里可以运用Python的 Shapely 库完成,Python 完成如下:
import numpy as np
import shapely
from shapely.errors import TopologicalError
from shapely.geometry import Polygon,MultiPoint
def to_polygon(quadrilateral):
"""
:param quadrilateral: 四边形四个点坐标的一维数组表示,[x,y,x,y....]
:return: 四边形二维数组, Polygon四边形对象
"""
# 四边形二维数组表示
quadrilateral_array = np.array(quadrilateral).reshape(4, 2)
# Polygon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
quadrilateral_polygon = Polygon(quadrilateral_array).convex_hull
return quadrilateral_array, quadrilateral_polygon
def calculate_iou(actual_quadrilateral, predict_quadrilateral):
"""
:param actual_quadrilateral: 预测四边形四个点坐标的一维数组表示,[x,y,x,y....]
:param predict_quadrilateral: 希冀四边形四个点坐标的一维数组表示,[x,y,x,y....]
:return:
"""
# 预测四边形二维数组, 预测四边形 Polygon 对象
actual_quadrilateral_array, actual_quadrilateral_polygon = to_polygon(actual_quadrilateral)
# 希冀四边形二维数组, 希冀四边形 Polygon 对象
predict_quadrilateral_array, predict_quadrilateral_polygon = to_polygon(predict_quadrilateral)
# 兼并两个box坐标,变为8*2 便于前面计算并集面积
union_poly = np.concatenate((actual_quadrilateral_array, predict_quadrilateral_array))
# 两两四边形能否存在交集
inter_status = actual_quadrilateral_polygon.intersects(predict_quadrilateral_polygon)
# 假设两四边形相交,则进iou计算
if inter_status:
try:
# 交集面积
inter_area = actual_quadrilateral_polygon.intersection(predict_quadrilateral_polygon).area
# 并集面积 计算方式一
#union_area = poly1.area + poly2.area - inter_area
# 并集面积 计算方式二
union_area = MultiPoint(union_poly).convex_hull.area
# 若并集面积等于0,则iou = 0
if union_area == 0:
iou = 0
else:
# 第一种计算的是: 交集部分/包含两个四边形最小多边形的面积
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