如何使用Shapely删除彼此太近的点？

Question

我得到了以下代码：

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)

现在我想编辑这样的点，如果两个相邻的点比MIN_DISTANCE更近，那么我想删除后一个:即从line=[(1, 1), (1.1, 1.1), (3, 3)], MIN_DISTANCE=2中我将获得line=[(1, 1), (3, 3)]。

是否可以编写一个强力解决方案(例如，for point in line)并覆盖直线上的点，或者是否有一个内置的函数？

Answer 1

您可以使用simplify方法来实现此目的：

new_line = line.simplify(tolerance=MIN_DISTANCE)

对于您的示例，它将是：

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]))