Python 如何使用Shapely删除彼此太近的点？

我有以下代码：

from shapely.geometry import LineString, Point
xs = range(10)
ys = range(10)
points = [Point(x, y) for x, y in zip(x, y)]
line = LineString(points)

现在我想编辑点，如果两个相邻点的距离小于

最小距离

，那么我想删除后一个点： i、 例如，从

line=[（1，1），（1.1，1.1），（3，3）]，最小距离=2

我会得到

line=[（1，1），（3，3）]

---

是否可以编写一个强力解决方案（即，

用于行内点

）并覆盖行内点，或者是否有内置函数？

您可以使用

简化

方法来实现：

new_line = line.simplify(tolerance=MIN_DISTANCE)

以您的示例为例：

from shapely.geometry import LineString, Point
xs = range(10)
ys = range(10)
x, y = zip(*[(1, 1), (1.1, 1.1), (3, 3)])
points = [Point(x, y) for x, y in zip(x, y)]
line = LineString(points)

MIN_DISTANCE = 2
new_line = line.simplify(tolerance=MIN_DISTANCE)

print(f"Original coordinates: {line.xy}")
print(f"Simplified coordinates: {new_line.xy}")


# Original coordinates: (array('d', [1.0, 1.1, 3.0]), array('d', [1.0, 1.1, 3.0]))
# Simplified coordinates: (array('d', [1.0, 3.0]), array('d', [1.0, 3.0]))

---

您可以使用

simplify

方法来实现：

new_line = line.simplify(tolerance=MIN_DISTANCE)

以您的示例为例：

from shapely.geometry import LineString, Point
xs = range(10)
ys = range(10)
x, y = zip(*[(1, 1), (1.1, 1.1), (3, 3)])
points = [Point(x, y) for x, y in zip(x, y)]
line = LineString(points)

MIN_DISTANCE = 2
new_line = line.simplify(tolerance=MIN_DISTANCE)

print(f"Original coordinates: {line.xy}")
print(f"Simplified coordinates: {new_line.xy}")


# Original coordinates: (array('d', [1.0, 1.1, 3.0]), array('d', [1.0, 1.1, 3.0]))
# Simplified coordinates: (array('d', [1.0, 3.0]), array('d', [1.0, 3.0]))