应该使用哪个 Shapely 谓词来区分这些 LinearRings

Question

在我的项目中有两个用例。在第一种情况下，红色和蓝色形状不重叠，但在第二种情况下却重叠。

案例#1

案例#2

在我的测试代码中，我尝试了来自 Shapely 的各种谓词，但无论如何都找不到区分它们的方法。

from shapely import LinearRing

# first use-case
blue_sq = LinearRing([(1, 1), (1, 2), (2, 2), (2, 1), (1, 1)])
red_one = LinearRing([(0, 0), (0, 3), (3, 3), (3, 2), (1, 2), (1, 0), (0, 0)])

print(blue_sq.overlaps(red_one))
print(blue_sq.intersects(red_one))
print(blue_sq.crosses(red_one))
print(blue_sq.contains(red_one))
print(blue_sq.touches(red_one))
print(blue_sq.within(red_one))

# second use-case
blue_ln = LinearRing([(2, 1), (2, 2), (7, 2), (7, 1), (2, 1)])
red_two = LinearRing([(1, 0), (1, 1), (2, 1), (2, 2), (3, 2), (3, 0), (1, 0)])

print(blue_ln.overlaps(red_two))
print(blue_ln.intersects(red_two))
print(blue_ln.crosses(red_two))
print(blue_ln.contains(red_two))
print(blue_ln.touches(red_two))
print(blue_ln.within(red_two))

结果总结于下表中。

形状谓词	案例1	案例2
重叠	正确	正确
相交	正确	正确
十字	错误	错误
包含	错误	错误
触动	错误	错误
内	错误	错误

两种情况看似交叉和重叠，但两种情况都没有交叉、包含、接触或位于对方之内。我确信我错过了一些明显的东西:-)

Answer 1

LinearRing 仍然只是一条线...所以考虑到这两种情况确实非常相似，并且谓词的结果没有差异。

当您将坐标视为曲面的外边界时，就会出现差异。要使用这样的线条，您需要使用它们来创建形状良好的多边形。如果你这样做，你会得到预期的不同结果。

代码示例：

from shapely import Polygon

# first use-case
blue_sq = Polygon([(1, 1), (1, 2), (2, 2), (2, 1), (1, 1)])
red_one = Polygon([(0, 0), (0, 3), (3, 3), (3, 2), (1, 2), (1, 0), (0, 0)])

print(f"{blue_sq.overlaps(red_one)=}")
print(f"{blue_sq.intersects(red_one)=}")
print(f"{blue_sq.crosses(red_one)=}")
print(f"{blue_sq.contains(red_one)=}")
print(f"{blue_sq.touches(red_one)=}")
print(f"{blue_sq.within(red_one)=}")

# second use-case
blue_ln = Polygon([(2, 1), (2, 2), (7, 2), (7, 1), (2, 1)])
red_two = Polygon([(1, 0), (1, 1), (2, 1), (2, 2), (3, 2), (3, 0), (1, 0)])

print(f"{blue_ln.overlaps(red_two)=}")
print(f"{blue_ln.intersects(red_two)=}")
print(f"{blue_ln.crosses(red_two)=}")
print(f"{blue_ln.contains(red_two)=}")
print(f"{blue_ln.touches(red_two)=}")
print(f"{blue_ln.within(red_two)=}")

输出：

blue_sq.overlaps(red_one)=False
blue_sq.intersects(red_one)=True
blue_sq.crosses(red_one)=False
blue_sq.contains(red_one)=False
blue_sq.touches(red_one)=True
blue_sq.within(red_one)=False
blue_ln.overlaps(red_two)=True
blue_ln.intersects(red_two)=True
blue_ln.crosses(red_two)=False
blue_ln.contains(red_two)=False
blue_ln.touches(red_two)=False
blue_ln.within(red_two)=False