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在目标检测当中，有一个重要的概念就是IOU。一般指代模型预测的bbox和Groud Truth之间的交并比。
图1：IoU
那么其计算公式就是
对IoU的计算也十分简单，如图2所示，一共有4种cases。
图2：cases
通常我们在代码中保存bbox有两种格式xyxy和xywh，这里我们使用xyxy更易于说明。假设两个框的bbox分别是$(x_1, y_1, x_2, y_2)$和$(x_3, y_3, x_4, y_4)$。那么相交的左上角顶点x坐标是$x_{lt}=max(x_1, x_3)$，y坐标是$y_{lt}=max(y_1, y_3)$；右下角顶点的x坐标是$x_{rb}=min(x_2, x_4)$，y坐标是$y_{rb}=max(y_2, y_4)$。这对于四种cases都是适用的。
那么相交的面积就可以计算出来为
$$ S_{intersection} = (y_{rb} - y_{lt}) \times (x_{rb} - x_{lt}) $$ 如果$y_{rb} - y_{lt}$或$x_{rb} - x_{lt}$有任何一个是负值，说明相交面积为0。两个bbox总面积为
$$ S_1 = (x_2 - x_1) \times (y_2 - y_1) \ S_2 = (x_4 - x_3) \times (y_4 - y_3) $$ 那么IoU的计算即为
IoF的计算和IoU相似，只不过分母是gt框的面积。
代码参考mmdetection中的IoU计算，讲解写在了代码中，不过代码中的原始注释还是比较详细的。
def bbox_overlaps(bboxes1, bboxes2, mode='iou', is_aligned=False, eps=1e-6):
"""Calculate overlap between two set of bboxes.
FP16 Contributed by https://github.com/open-mmlab/mmdetection/pull/4889
Note:
Assume bboxes1 is M x 4, bboxes2 is N x 4, when mode is 'iou',
there are some new generated variable when calculating IOU
using bbox_overlaps function:
1) is_aligned is False
area1: M x 1
area2: N x 1
lt: M x N x 2
rb: M x N x 2
wh: M x N x 2
overlap: M x N x 1
union: M x N x 1
ious: M x N x 1
Total memory:
S = (9 x N x M + N + M) * 4 Byte,
When using FP16, we can reduce:
R = (9 x N x M + N + M) * 4 / 2 Byte
R large than (N + M) * 4 * 2 is always true when N and M >= 1.
Obviously, N + M <= N * M < 3 * N * M, when N >=2 and M >=2,
N + 1 < 3 * N, when N or M is 1.
Given M = 40 (ground truth), N = 400000 (three anchor boxes
in per grid, FPN, R-CNNs),
R = 275 MB (one times)
A special case (dense detection), M = 512 (ground truth),
R = 3516 MB = 3.43 GB
When the batch size is B, reduce:
B x R
Therefore, CUDA memory runs out frequently.
Experiments on GeForce RTX 2080Ti (11019 MiB):
| dtype | M | N | Use | Real | Ideal |
|:----:|:----:|:----:|:----:|:----:|:----:|
| FP32 | 512 | 400000 | 8020 MiB | -- | -- |
| FP16 | 512 | 400000 | 4504 MiB | 3516 MiB | 3516 MiB |
| FP32 | 40 | 400000 | 1540 MiB | -- | -- |
| FP16 | 40 | 400000 | 1264 MiB | 276MiB | 275 MiB |
2) is_aligned is True
area1: N x 1
area2: N x 1
lt: N x 2
rb: N x 2
wh: N x 2
overlap: N x 1
union: N x 1
ious: N x 1
Total memory:
S = 11 x N * 4 Byte
When using FP16, we can reduce:
R = 11 x N * 4 / 2 Byte
So do the 'giou' (large than 'iou').
Time-wise, FP16 is generally faster than FP32.
When gpu_assign_thr is not -1, it takes more time on cpu
but not reduce memory.
There, we can reduce half the memory and keep the speed.
If ``is_aligned`` is ``False``, then calculate the overlaps between each
bbox of bboxes1 and bboxes2, otherwise the overlaps between each aligned
pair of bboxes1 and bboxes2.
Args:
bboxes1 (Tensor): shape (B, m, 4) in <x1, y1, x2, y2> format or empty.
bboxes2 (Tensor): shape (B, n, 4) in <x1, y1, x2, y2> format or empty.
B indicates the batch dim, in shape (B1, B2, ..., Bn).
If ``is_aligned`` is ``True``, then m and n must be equal.
mode (str): "iou" (intersection over union), "iof" (intersection over
foreground) or "giou" (generalized intersection over union).
Default "iou".
is_aligned (bool, optional): If True, then m and n must be equal.
Default False.
eps (float, optional): A value added to the denominator for numerical
stability. Default 1e-6.
Returns:
Tensor: shape (m, n) if ``is_aligned`` is False else shape (m,)
Example:
>>> bboxes1 = torch.FloatTensor([
>>> [0, 0, 10, 10],
>>> [10, 10, 20, 20],
>>> [32, 32, 38, 42],
>>> ])
>>> bboxes2 = torch.FloatTensor([
>>> [0, 0, 10, 20],
>>> [0, 10, 10, 19],
>>> [10, 10, 20, 20],
>>> ])
>>> overlaps = bbox_overlaps(bboxes1, bboxes2)
>>> assert overlaps.shape == (3, 3)
>>> overlaps = bbox_overlaps(bboxes1, bboxes2, is_aligned=True)
>>> assert overlaps.shape == (3, )
Example:
>>> empty = torch.empty(0, 4)
>>> nonempty = torch.FloatTensor([[0, 0, 10, 9]])
>>> assert tuple(bbox_overlaps(empty, nonempty).shape) == (0, 1)
>>> assert tuple(bbox_overlaps(nonempty, empty).shape) == (1, 0)
>>> assert tuple(bbox_overlaps(empty, empty).shape) == (0, 0)
"""
assert mode in ['iou', 'iof', 'giou'], f'Unsupported mode {mode}'
# Either the boxes are empty or the length of boxes' last dimension is 4
assert (bboxes1.size(-1) == 4 or bboxes1.size(0) == 0)
assert (bboxes2.size(-1) == 4 or bboxes2.size(0) == 0)
# Batch dim must be the same
# Batch dim: (B1, B2, ... Bn)
assert bboxes1.shape[:-2] == bboxes2.shape[:-2]
batch_shape = bboxes1.shape[:-2]
rows = bboxes1.size(-2)
cols = bboxes2.size(-2)
if is_aligned:
assert rows == cols
if rows * cols == 0:
if is_aligned:
return bboxes1.new(batch_shape + (rows, ))
else:
return bboxes1.new(batch_shape + (rows, cols))
area1 = (bboxes1[..., 2] - bboxes1[..., 0]) * (
bboxes1[..., 3] - bboxes1[..., 1])
area2 = (bboxes2[..., 2] - bboxes2[..., 0]) * (
bboxes2[..., 3] - bboxes2[..., 1])
if is_aligned:
lt = torch.max(bboxes1[..., :2], bboxes2[..., :2]) # [B, rows, 2]
rb = torch.min(bboxes1[..., 2:], bboxes2[..., 2:]) # [B, rows, 2]
wh = fp16_clamp(rb - lt, min=0)
overlap = wh[..., 0] * wh[..., 1]
if mode in ['iou', 'giou']:
union = area1 + area2 - overlap
else:
union = area1
if mode == 'giou':
enclosed_lt = torch.min(bboxes1[..., :2], bboxes2[..., :2])
enclosed_rb = torch.max(bboxes1[..., 2:], bboxes2[..., 2:])
else:
lt = torch.max(bboxes1[..., :, None, :2],
bboxes2[..., None, :, :2]) # [B, rows, cols, 2]
rb = torch.min(bboxes1[..., :, None, 2:],
bboxes2[..., None, :, 2:]) # [B, rows, cols, 2]
wh = fp16_clamp(rb - lt, min=0)
overlap = wh[..., 0] * wh[..., 1]
if mode in ['iou', 'giou']:
union = area1[..., None] + area2[..., None, :] - overlap
else:
union = area1[..., None]
if mode == 'giou':
enclosed_lt = torch.min(bboxes1[..., :, None, :2],
bboxes2[..., None, :, :2])
enclosed_rb = torch.max(bboxes1[..., :, None, 2:],
bboxes2[..., None, :, 2:])
eps = union.new_tensor([eps])
union = torch.max(union, eps)
ious = overlap / union
if mode in ['iou', 'iof']:
return ious
# calculate gious
enclose_wh = fp16_clamp(enclosed_rb - enclosed_lt, min=0)
enclose_area = enclose_wh[..., 0] * enclose_wh[..., 1]
enclose_area = torch.max(enclose_area, eps)
gious = ious - (enclose_area - union) / enclose_area
return gious