主要用来求RoI和GT的IoU, 也可以用来求预测结果和GT的IoU
def iou(boxes1, boxes2):
"""Computes IoU overlaps between two sets of boxes.
boxes1, boxes2: [N, (y1, x1, y2, x2)].
"""
# 1. Tile boxes2 and repeat boxes1. This allows us to compare
# every boxes1 against every boxes2 without loops.
# TF doesn't have an equivalent to np.repeat() so simulate it
# using tf.tile() and tf.reshape().
b1 = np.repeat(boxes1, np.shape(boxes2)[0], axis=0)
b2 = np.tile(boxes2, [np.shape(boxes1)[0], 1])
# 2. Compute intersections
b1_y1, b1_x1, b1_y2, b1_x2 = np.split(b1, 4, axis=1)
b2_y1, b2_x1, b2_y2, b2_x2 = np.split(b2, 4, axis=1)
y1 = np.maximum(b1_y1, b2_y1)
x1 = np.maximum(b1_x1, b2_x1)
y2 = np.minimum(b1_y2, b2_y2)
x2 = np.minimum(b1_x2, b2_x2)
intersection = np.maximum(x2 - x1, 0) * np.maximum(y2 - y1, 0)
# 3. Compute unions
b1_area = (b1_y2 - b1_y1) * (b1_x2 - b1_x1)
b2_area = (b2_y2 - b2_y1) * (b2_x2 - b2_x1)
union = b1_area + b2_area - intersection
# 4. Compute IoU and reshape to [boxes1, boxes2]
iou = intersection / union
overlaps = np.reshape(iou, [np.shape(boxes1)[0], np.shape(boxes2)[0]])
return overlaps
roi_iou_max = tf.reduce_max(overlaps, axis=1)
第一步即先把二者形状统一以方便逐个比较:
boxes1 = [[0,0,3,3], [1,1,5,5], [9,9,11,11]]
boxes2 = [[0,0,2,2], [1,1,3,3], [4,4,6,6], [6,6,9,9], [11,11,13,13]]
输出为
[[ 0 0 3 3]
[ 0 0 3 3]
[ 0 0 3 3]
[ 0 0 3 3]
[ 0 0 3 3]
[ 1 1 5 5]
[ 1 1 5 5]
[ 1 1 5 5]
[ 1 1 5 5]
[ 1 1 5 5]
[ 9 9 11 11]
[ 9 9 11 11]
[ 9 9 11 11]
[ 9 9 11 11]
[ 9 9 11 11]]
[[ 0 0 2 2]
[ 1 1 3 3]
[ 4 4 6 6]
[ 6 6 9 9]
[11 11 13 13]
[ 0 0 2 2]
[ 1 1 3 3]
[ 4 4 6 6]
[ 6 6 9 9]
[11 11 13 13]
[ 0 0 2 2]
[ 1 1 3 3]
[ 4 4 6 6]
[ 6 6 9 9]
[11 11 13 13]]
之后的交集, 并集, IoU都是15x1的矩阵, 最后reshape成3x5的即[np.shape(boxes1)[0], np.shape(boxes2)[0]]代表把boxes1的每一项逐个和boxes2的每一项比较IoU, boxes1中同一个box得到的结果在同一行.
