使用原始点列表在 Python/cv2 中查找重叠矩形的区域

Question

我有一个矩形的坐标x1, y1, x2,y2和其他矩形的其他坐标列表。

我想将我已经拥有的值与其他值进行比较，以查看它们是否比50%原始矩形重叠更多。

我检查了其他资源，但仍然可以理解它：

• https://codereview.stackexchange.com/questions/151309/check-if-two-rectangles-overlap
• https://codereview.stackexchange.com/questions/31352/overlapping-rectangles

Answer 1

让我们首先将您的问题简化为单一维度：

你有一个区间A = [a0, a1]，想知道另一个区间B = [b0, b1]与它相交多少。我将代表A与=和B用-。

有6种可能的情况：

• A包含B,intersection = b1 - b0

  a0  b0  b1  a1
  ==============
      ------
  

• B包含A,intersection = a1 - a0

  b0  a0  a1  b1
      ======
  --------------
  

• BA从左边相交，intersection = b1 - a0

  b0  a0  b1  a1
      ==========
  ----------
  

• BA从右边相交，intersection = a1 - b0

  a0  b0  a1  b1
  ==========
      ----------
  

• B位于 的左侧A，intersection = 0

  b0  b1  a0  a1
          ======
  ------
  

• B在 的右边A，intersection = 0

  a0  a1  b0  b1
  ======
          ------
  

基于此，我们可以定义一个函数，在给定两个区间A = [a0, a1]和 的情况下B = [b0, b1]，返回它们相交的程度：

def calculateIntersection(a0, a1, b0, b1):
    if a0 >= b0 and a1 <= b1: # Contained
        intersection = a1 - a0
    elif a0 < b0 and a1 > b1: # Contains
        intersection = b1 - b0
    elif a0 < b0 and a1 > b0: # Intersects right
        intersection = a1 - b0
    elif a1 > b1 and a0 < b1: # Intersects left
        intersection = b1 - a0
    else: # No intersection (either side)
        intersection = 0

    return intersection

这几乎就是你需要的一切。要找出两个矩形相交多少，您只需要在X和Y轴上执行此函数并将这些数量相乘以获得相交区域：

# The rectangle against which you are going to test the rest and its area:
X0, Y0, X1, Y1, = [0, 0, 10, 10]
AREA = float((X1 - X0) * (Y1 - Y0))

# Rectangles to check
rectangles = [[15, 0, 20, 10], [0, 15, 10, 20], [0, 0, 5, 5], [0, 0, 5, 10], [0, 5, 10, 100], [0, 0, 100, 100]]

# Intersecting rectangles:
intersecting = []

for x0, y0, x1, y1 in rectangles:       
    width = calculateIntersection(x0, x1, X0, X1)        
    height = calculateIntersection(y0, y1, Y0, Y1)        
    area = width * height
    percent = area / AREA

    if (percent >= 0.5):
        intersecting.append([x0, y0, x1, y1])

结果将是：

• [15, 0, 20, 10]在 X 轴上不相交，所以width = 0.
• [0, 15, 10, 20]在 Y 轴上不相交，所以height = 0.
• [0, 0, 5, 5]只相交25%。
• [0, 0, 5, 10]相交50%并将被添加到intersecting.
• [0, 5, 10, 100]相交50%并将被添加到intersecting.
• [0, 0, 100, 100]相交100%并将被添加到intersecting.