38

有一条折线，其中包含顶点坐标列表 = [(x1,y1), (x2,y2), (x3,y3),...] 和一个点 (x,y)。在 Shapely 中，geometry1.distance(geometry2)返回两个几何图形之间的最短距离。

>>> from shapely.geometry import LineString, Point
>>> line = LineString([(0, 0), (5, 7), (12, 6)])  # geometry2
>>> list(line.coords)
[(0.0, 0.0), (5.0, 7.0), (12.0, 6.0)]
>>> p = Point(4,8)  # geometry1
>>> list(p.coords)
[(4.0, 8.0)]
>>> p.distance(line)
1.4142135623730951

但我还需要找到最接近点（x，y）的线上点的坐标。在上面的示例中，这是LineString对象上距离 1.4142135623730951 单位的点的坐标Point(4,8)。该方法distance()在计算距离时应该有坐标。有没有办法从这个方法返回它？

4

2 回答 2

55

您描述的 GIS 术语是线性参考，Shapely 有这些方法。

# Length along line that is closest to the point
print(line.project(p))

# Now combine with interpolated point on line
p2 = line.interpolate(line.project(p))
print(p2)  # POINT (5 7)

另一种方法是使用nearest_points：

from shapely.ops import nearest_points
p2 = nearest_points(line, p)[0]
print(p2)  # POINT (5 7)

它提供了与线性参考技术相同的答案，但可以从更复杂的几何输入（如两个多边形）中确定最近的一对点。

于 2014-06-26T20:58:23.833 回答
3

如果您有一个段（例如：一行，指的是标题）而不是段列表，这就是我所做的，并且通过了一个测试用例。请注意，此页面上的某些用户只是通过查看标题来寻找来自 Google 搜索的内容。

Python代码：

def sq_shortest_dist_to_point(self, other_point):
    dx = self.b.x - self.a.x
    dy = self.b.y - self.a.y
    dr2 = float(dx ** 2 + dy ** 2)

    lerp = ((other_point.x - self.a.x) * dx + (other_point.y - self.a.y) * dy) / dr2
    if lerp < 0:
        lerp = 0
    elif lerp > 1:
        lerp = 1

    x = lerp * dx + self.a.x
    y = lerp * dy + self.a.y

    _dx = x - other_point.x
    _dy = y - other_point.y
    square_dist = _dx ** 2 + _dy ** 2
    return square_dist

def shortest_dist_to_point(self, other_point):
    return math.sqrt(self.sq_shortest_dist_to_point(other_point))

一个测试用例：

def test_distance_to_other_point(self):
    # Parametrize test with multiple cases:
    segments_and_point_and_answer = [
        [Segment(Point(1.0, 1.0), Point(1.0, 3.0)), Point(2.0, 4.0), math.sqrt(2.0)],
        [Segment(Point(1.0, 1.0), Point(1.0, 3.0)), Point(2.0, 3.0), 1.0],
        [Segment(Point(0.0, 0.0), Point(0.0, 3.0)), Point(1.0, 1.0), 1.0],
        [Segment(Point(1.0, 1.0), Point(3.0, 3.0)), Point(2.0, 2.0), 0.0],
        [Segment(Point(-1.0, -1.0), Point(3.0, 3.0)), Point(2.0, 2.0), 0.0],
        [Segment(Point(1.0, 1.0), Point(1.0, 3.0)), Point(2.0, 3.0), 1.0],
        [Segment(Point(1.0, 1.0), Point(1.0, 3.0)), Point(2.0, 4.0), math.sqrt(2.0)],
        [Segment(Point(1.0, 1.0), Point(-3.0, -3.0)), Point(-3.0, -4.0), 1],
        [Segment(Point(1.0, 1.0), Point(-3.0, -3.0)), Point(-4.0, -3.0), 1],
        [Segment(Point(1.0, 1.0), Point(-3.0, -3.0)), Point(1, 2), 1],
        [Segment(Point(1.0, 1.0), Point(-3.0, -3.0)), Point(2, 1), 1],
        [Segment(Point(1.0, 1.0), Point(-3.0, -3.0)), Point(-3, -1), math.sqrt(2.0)],
        [Segment(Point(1.0, 1.0), Point(-3.0, -3.0)), Point(-1, -3), math.sqrt(2.0)],
        [Segment(Point(-1.0, -1.0), Point(3.0, 3.0)), Point(3, 1), math.sqrt(2.0)],
        [Segment(Point(-1.0, -1.0), Point(3.0, 3.0)), Point(1, 3), math.sqrt(2.0)],
        [Segment(Point(1.0, 1.0), Point(3.0, 3.0)), Point(3, 1), math.sqrt(2.0)],
        [Segment(Point(1.0, 1.0), Point(3.0, 3.0)), Point(1, 3), math.sqrt(2.0)]
    ]

    for i, (segment, point, answer) in enumerate(segments_and_point_and_answer):
        result = segment.shortest_dist_to_point(point)
        self.assertAlmostEqual(result, answer, delta=0.001, msg=str((i, segment, point, answer)))

注意：我假设这个函数在一个Segment类中。如果您的线是无限的，请不要lerp仅将 0 限制为 1，但仍至少提供两个不同的a和b点。

于 2018-03-27T04:20:43.367 回答