我有以下问题。我尝试计算并集的交集,即两个分量的重叠除以两个分量的并集。假设组件 1 是一个矩阵,其中第一个对象所在的位置,组件 2 是一个矩阵,其中第二个对象所在的位置。我可以用
np.logical_and(component == 1, component2 == 1)
来计算重叠。但我如何计算联盟呢?我只对相连的物体感兴趣。
import numpy as np
component1 = np.array([[0,1,1],[0,1,1],[0,1,1]])
component2 = np.array([[1,1,0],[1,1,0],[1,1,0]])
overlap = np.logical_and(component == 1, component2 == 1)
union = ?
IOU = len(overlap)/len(union)
如果您只处理
0
和 1
,那么使用布尔数组会更容易:
import numpy as np
component1 = np.array([[0,1,1],[0,1,1],[0,1,1]], dtype=bool)
component2 = np.array([[1,1,0],[1,1,0],[1,1,0]], dtype=bool)
overlap = component1*component2 # Logical AND
union = component1 + component2 # Logical OR
IOU = overlap.sum()/float(union.sum()) # Treats "True" as 1,
# sums number of Trues
# in overlap and union
# and divides
>>> 1*overlap
array([[0, 1, 0],
[0, 1, 0],
[0, 1, 0]])
>>> 1*union
array([[1, 1, 1],
[1, 1, 1],
[1, 1, 1]])
>>> IOU
0.3333333333333333
Oleksii 在 medium 中为您的问题提供了答案。
简单来说:
intersection = numpy.logical_and(result1, result2)
union = numpy.logical_or(result1, result2)
iou_score = numpy.sum(intersection) / numpy.sum(union)
print(‘IoU is %s’ % iou_score)
此外,他对此也给出了很好的解释。看看上面的链接。
只需使用专用的 numpy 函数
intersect1d
和 union1d
:
intersection = np.intersect1d(a, b)
union = np.union1d(a, b)
iou = intersection.shape[0] / union.shape[0]
这么多变体 - 我将添加另一个:
import numpy as np
SMOOTH = 1e-6
def iou_numpy(outputs: np.array, labels: np.array):
outputs = outputs.squeeze(1)
intersection = (outputs & labels).sum((1, 2))
union = (outputs | labels).sum((1, 2))
iou = (intersection + SMOOTH) / (union + SMOOTH)
thresholded = np.ceil(np.clip(20 * (iou - 0.7), 0, 10)) / 10
return thresholded # Or thresholded.mean()
我在这里得到了这段代码 - 我真的很喜欢这个页面如何解释 IoU 指标的概念。