Python:交集与并集

Question

我有以下问题。我尝试计算Union上的交集，它是两个组件的重叠除以两个组件的单位单位。让我们假设component1是一个矩阵，其中第一个对象是1，component2是一个矩阵，第二个对象是1。我可以用np.logical_and(component == 1, component2 == 1)计算重叠部分。但是我怎么才能计算出工会呢？我只对连接的对象感兴趣。

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)

Answer 1

如果您只处理0和1，则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

Answer 2

Oleksii在medium中为您的问题提供了答案。

简单地说：

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，他对此做了很好的解释。看看上面的链接。

Answer 3

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]