Question

我有两个2D旋转矩形，定义为（中心x，中心y，高度，宽度）和旋转角度（0-360°）。我如何计算这两个旋转矩形的交叉区域。

Answer 1

使用计算几何包，例如， Shapely：

import shapely.geometry
import shapely.affinity

class RotatedRect:
    def __init__(self, cx, cy, w, h, angle):
        self.cx = cx
        self.cy = cy
        self.w = w
        self.h = h
        self.angle = angle

    def get_contour(self):
        w = self.w
        h = self.h
        c = shapely.geometry.box(-w/2.0, -h/2.0, w/2.0, h/2.0)
        rc = shapely.affinity.rotate(c, self.angle)
        return shapely.affinity.translate(rc, self.cx, self.cy)

    def intersection(self, other):
        return self.get_contour().intersection(other.get_contour())


r1 = RotatedRect(10, 15, 15, 10, 30)
r2 = RotatedRect(15, 15, 20, 10, 0)

from matplotlib import pyplot
from descartes import PolygonPatch

fig = pyplot.figure(1, figsize=(10, 4))
ax = fig.add_subplot(121)
ax.set_xlim(0, 30)
ax.set_ylim(0, 30)

ax.add_patch(PolygonPatch(r1.get_contour(), fc='#990000', alpha=0.7))
ax.add_patch(PolygonPatch(r2.get_contour(), fc='#000099', alpha=0.7))
ax.add_patch(PolygonPatch(r1.intersection(r2), fc='#009900', alpha=1))

pyplot.show()

Answer 2

这是一个不使用Python标准库之外的任何库的解决方案。

确定两个矩形的交叉区域可分为两个子问题：

查找交叉点多边形（如果有）;
确定交叉点多边形的区域。

使用时，这两个问题都比较容易矩形的顶点（角）。所以首先你必须确定这些顶点。假设坐标原点位于中心矩形，顶点是，从逆时针方向的左下角开始： (-w/2, -h/2)，(w/2, -h/2)，(w/2, h/2)和(-w/2, h/2)。在角度a上旋转，并翻译它们到矩形中心的正确位置，这些变为： (cx + (-w/2)cos(a) - (-h/2)sin(a), cy + (-w/2)sin(a) + (-h/2)cos(a))，其他角点类似。

确定交叉点多边形的一种简单方法如下：从一个矩形开始作为候选交叉多边形。然后应用顺序切割过程（如here所述。简而言之：你依次取第二个矩形的每个边缘，并从＆＃34;外部＆＃34;上的候选交叉点多边形中删除所有部分。由边缘定义的半平面（向两个方向延伸）。对所有边执行此操作会留下候选交叉点多边形仅包含第二个矩形内或其边界内的部分。

可以计算得到的多边形的面积（由一系列顶点定义）从顶点的坐标。您总结顶点的叉积每个边缘（再次按逆时针顺序），并将其除以2。参见例如www.mathopenref.com/coordpolygonarea.html

足够的理论和解释。这是代码：

from math import pi, cos, sin


class Vector:
    def __init__(self, x, y):
        self.x = x
        self.y = y

    def __add__(self, v):
        if not isinstance(v, Vector):
            return NotImplemented
        return Vector(self.x + v.x, self.y + v.y)

    def __sub__(self, v):
        if not isinstance(v, Vector):
            return NotImplemented
        return Vector(self.x - v.x, self.y - v.y)

    def cross(self, v):
        if not isinstance(v, Vector):
            return NotImplemented
        return self.x*v.y - self.y*v.x


class Line:
    # ax + by + c = 0
    def __init__(self, v1, v2):
        self.a = v2.y - v1.y
        self.b = v1.x - v2.x
        self.c = v2.cross(v1)

    def __call__(self, p):
        return self.a*p.x + self.b*p.y + self.c

    def intersection(self, other):
        # See e.g.     https://en.wikipedia.org/wiki/Line%E2%80%93line_intersection#Using_homogeneous_coordinates
        if not isinstance(other, Line):
            return NotImplemented
        w = self.a*other.b - self.b*other.a
        return Vector(
            (self.b*other.c - self.c*other.b)/w,
            (self.c*other.a - self.a*other.c)/w
        )


def rectangle_vertices(cx, cy, w, h, r):
    angle = pi*r/180
    dx = w/2
    dy = h/2
    dxcos = dx*cos(angle)
    dxsin = dx*sin(angle)
    dycos = dy*cos(angle)
    dysin = dy*sin(angle)
    return (
        Vector(cx, cy) + Vector(-dxcos - -dysin, -dxsin + -dycos),
        Vector(cx, cy) + Vector( dxcos - -dysin,  dxsin + -dycos),
        Vector(cx, cy) + Vector( dxcos -  dysin,  dxsin +  dycos),
        Vector(cx, cy) + Vector(-dxcos -  dysin, -dxsin +  dycos)
    )

def intersection_area(r1, r2):
    # r1 and r2 are in (center, width, height, rotation) representation
    # First convert these into a sequence of vertices

    rect1 = rectangle_vertices(*r1)
    rect2 = rectangle_vertices(*r2)

    # Use the vertices of the first rectangle as
    # starting vertices of the intersection polygon.
    intersection = rect1

    # Loop over the edges of the second rectangle
    for p, q in zip(rect2, rect2[1:] + rect2[:1]):
        if len(intersection) <= 2:
            break # No intersection

        line = Line(p, q)

        # Any point p with line(p) <= 0 is on the "inside" (or on the boundary),
        # any point p with line(p) > 0 is on the "outside".

        # Loop over the edges of the intersection polygon,
        # and determine which part is inside and which is outside.
        new_intersection = []
        line_values = [line(t) for t in intersection]
        for s, t, s_value, t_value in zip(
            intersection, intersection[1:] + intersection[:1],
            line_values, line_values[1:] + line_values[:1]):
            if s_value <= 0:
                new_intersection.append(s)
            if s_value * t_value < 0:
                # Points are on opposite sides.
                # Add the intersection of the lines to new_intersection.
                intersection_point = line.intersection(Line(s, t))
                new_intersection.append(intersection_point)

        intersection = new_intersection

    # Calculate area
    if len(intersection) <= 2:
        return 0

    return 0.5 * sum(p.x*q.y - p.y*q.x for p, q in
                     zip(intersection, intersection[1:] + intersection[:1]))


if __name__ == '__main__':
    r1 = (10, 15, 15, 10, 30)
    r2 = (15, 15, 20, 10, 0)
    print(intersection_area(r1, r2))

Answer 3

"".contains("")       // true
"".contains("")       // true
"‍‍‍".contains("‍‍‍")      // true
"‍‍‍".contains("")      // true. In swift 3, this prints false
"‍‍‍".contains("\u{200D}") // false
"‍‍‍".contains("")      // true. In swift 3, this prints false
"‍‍‍".contains("")      // true

在查看此问题的可能重复页面后，我找不到完整的python答案，所以这是我使用屏蔽的解决方案。此功能适用于任何角度的复杂形状，而不仅仅是矩形

您将旋转的矩形的2个轮廓作为参数传递，如果没有发生交叉，则返回“无”，或者相交于原始图像的相交区域的图像和该图像的左/上位置取自

使用python，cv2和numpy

intersection, pnt = contourIntersection(rect1, rect2)

要完成答案，您可以使用上述功能，使用2个旋转矩形的CenterX，CenterY，Width，Height和Angle的值，我添加了以下功能。简单地将代码底部的Rect1和Rect2属性更改为您自己的

import cv2
import math
import numpy as np


def contourIntersection(con1, con2, showContours=False):

    # skip if no bounding rect intersection
    leftmost1 = tuple(con1[con1[:, :, 0].argmin()][0])
    topmost1 = tuple(con1[con1[:, :, 1].argmin()][0])
    leftmost2 = tuple(con2[con2[:, :, 0].argmin()][0])
    topmost2 = tuple(con2[con2[:, :, 1].argmin()][0])

    rightmost1 = tuple(con1[con1[:, :, 0].argmax()][0])
    bottommost1 = tuple(con1[con1[:, :, 1].argmax()][0])
    rightmost2 = tuple(con2[con2[:, :, 0].argmax()][0])
    bottommost2 = tuple(con2[con2[:, :, 1].argmax()][0])

    if rightmost1[0] < leftmost2[0] or rightmost2[0] < leftmost1[0] or bottommost1[1] < topmost2[1] or bottommost2[1] < topmost1[1]:
        return None, None

    # reset top / left to 0
    left = leftmost1[0] if leftmost1[0] < leftmost2[0] else leftmost2[0]
    top = topmost1[1] if topmost1[1] < topmost2[1] else topmost2[1]

    newCon1 = []
    for pnt in con1:

        newLeft = pnt[0][0] - left
        newTop = pnt[0][1] - top

        newCon1.append([newLeft, newTop])
    # next
    con1_new = np.array([newCon1], dtype=np.int32)

    newCon2 = []
    for pnt in con2:

        newLeft = pnt[0][0] - left
        newTop = pnt[0][1] - top

        newCon2.append([newLeft, newTop])
    # next
    con2_new = np.array([newCon2], dtype=np.int32)

    # width / height
    right1 = rightmost1[0] - left
    bottom1 = bottommost1[1] - top
    right2 = rightmost2[0] - left
    bottom2 = bottommost2[1] - top

    width = right1 if right1 > right2 else right2
    height = bottom1 if bottom1 > bottom2 else bottom2

    # create images
    img1 = np.zeros([height, width], np.uint8)
    cv2.drawContours(img1, con1_new, -1, (255, 255, 255), -1)

    img2 = np.zeros([height, width], np.uint8)
    cv2.drawContours(img2, con2_new, -1, (255, 255, 255), -1)

    # mask images together using AND
    imgIntersection = cv2.bitwise_and(img1, img2)

    if showContours:
        img1[img1 > 254] = 128
        img2[img2 > 254] = 100

        imgAll = cv2.bitwise_or(img1, img2)
        cv2.imshow('Merged Images', imgAll)

    # end if

    if not imgIntersection.sum():
        return None, None

    # trim
    while not imgIntersection[0].sum():
        imgIntersection = np.delete(imgIntersection, (0), axis=0)
        top += 1

    while not imgIntersection[-1].sum():
        imgIntersection = np.delete(imgIntersection, (-1), axis=0)

    while not imgIntersection[:, 0].sum():
        imgIntersection = np.delete(imgIntersection, (0), axis=1)
        left += 1

    while not imgIntersection[:, -1].sum():
        imgIntersection = np.delete(imgIntersection, (-1), axis=1)

    return imgIntersection, (left, top)
# end function

计算python中两个旋转矩形的交集区域

3 个答案: