如何在Shapely中获得线上等距点

我试图（大致）将一条线的点均匀地间隔到预定义的距离。

距离之间有一定的公差是可以的，但最好尽可能接近。

我知道我可以手动迭代线路中的每个点并检查 p1 与 p2 的距离，并根据需要添加更多点。

但我想知道是否有人知道是否有办法用 shapely 来实现这一点，因为我已经在 LineString 中拥有了坐标。

一种方法是使用interpolate返回沿线指定距离处的点的方法。您只需首先以某种方式生成距离列表。以输入行为例罗伊2012年的回答:

import numpy as np
from shapely.geometry import LineString
from shapely.ops import unary_union

line = LineString(([0, 0], [2, 1], [3, 2], [3.5, 1], [5, 2]))

以指定距离分裂：

distance_delta = 0.9
distances = np.arange(0, line.length, distance_delta)
# or alternatively without NumPy:
# points_count = int(line.length // distance_delta) + 1
# distances = (distance_delta * i for i in range(points_count))
points = [line.interpolate(distance) for distance in distances] + [line.boundary[1]]
multipoint = unary_union(points)  # or new_line = LineString(points)

Note that since the distance is fixed you can have problems at the end of the line as shown in the image. Depending on what you want you can include/exclude the [line.boundary[1]] part which adds the line's endpoint or use distances = np.arange(0, line.length, distance_delta)[:-1] to exclude the penultimate point.

另请注意，unary_union我正在使用应该比调用更有效object.union(other)在循环内，如另一个答案所示。

分割为固定数量的点：

n = 7
# or to get the distances closest to the desired one:
# n = round(line.length / desired_distance_delta)
distances = np.linspace(0, line.length, n)
# or alternatively without NumPy:
# distances = (line.length * i / (n - 1) for i in range(n))
points = [line.interpolate(distance) for distance in distances]
multipoint = unary_union(points)  # or new_line = LineString(points)