首页 > 解决方案 > 如何将 LineString 拆分为段

问题描述

我的数据集由 a 组成LineString,我想过滤掉它的各个线段LineString。更准确地说,每一个街道段。

到目前为止,我已经从数据集中提取了各个点并将它们保存在单独的列表中。此外,我想再次收集这些点并从中创建单独的 LineStrings 以将它们存储到 Geodataframe 中。数据有这种形式:

LINESTRING (3275.284016199762 340555.8579582386, 3241.504528076811 340504.1348617533, 3245.415803206172 340501.457084205, 3280.414559049542 340552.7138220053, 3285.19053022

我的问题是我必须LineString为每次迭代创建并显式保存一个单独的。谁能帮我这个?有没有更好的方法呢?

from shapely.geometry import Point, LineString

#Loop over LineString and gather Points
c=[]

for i in range(0,end):
    c.append(Point(route1.coords[i]))


iterator=len(c)
max=len(c)-1

#Loop to store LineStrings - got stuck here
for i in np.arange(0,iterator):
    if i<max:
        LineString([c[i], c[i+1]]).wkt

    else:
        break;

输出应如下所示:

Linestring(Point A, Point B)  
Linestring(Point B, Point C)  
Linestring(Point C, Point D)  
...  
Linestring(Point Y, Point Z)

标签: pythonpostgisgeopandasshapely

解决方案


说到 Shapely,它没有提供将曲线对象(LineStringLinearRing)分割成线段的功能,因此您必须自己编写。这是一个示例,说明如何使用zip迭代坐标对并将map它们迭代到LineString's:

from shapely.geometry import LineString, LinearRing


def segments(curve):
    return list(map(LineString, zip(curve.coords[:-1], curve.coords[1:])))


line = LineString([(0, 0), (1, 1), (2, 2)])
ring = LinearRing([(0, 0), (1, 0), (1, 1), (0, 1)])

line_segments = segments(line)
for segment in line_segments:
    print(segment)
# LINESTRING (0 0, 1 1)
# LINESTRING (1 1, 2 2)

ring_segments = segments(ring)
for segment in ring_segments:
    print(segment)
# LINESTRING (0 0, 1 0)
# LINESTRING (1 0, 1 1)
# LINESTRING (1 1, 0 1)
# LINESTRING (0 1, 0 0)

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