How to get the endpoint of a LineString in Shapely

Question

Linestring1 = LINESTRING (51.2176008 4.4177154, 51.21758 4.4178548, **51.2175729 4.4179023**, *51.21745162000732 4.41871738126533*)
Linestring2 = LINESTRING (*51.21745162000732 4.41871738126533*, **51.2174025 4.4190475**, 51.217338 4.4194807, 51.2172511 4.4200562, 51.2172411 4.4201077, 51.2172246 4.4201654, 51.2172067 4.420205, 51.2171806 4.4202355, 51.2171074 4.4202929, 51.2170063 4.4203409, 51.2169564 4.4203641, 51.2168076 4.4204243, 51.2166588 4.4204833, 51.2159018 4.420431, 51.2154117 4.4203843)

Considering these two linestrings were cut from a bigger linestring, how to get the endpoint of a LineString?

- Point(51.21745162000732 4.41871738126533) removed

- The new last element of linestring 1 = “ 51.2175729 4.4179023

- The new first element of linestring 2 = “ 51.2174025 4.4190475

In short, I want to get the new last value of the first part (linestring1) and the new first value of the second part (linestring2), but without the point where I cut them. How can I make this work?

Iconic

Answer 1

To get endpoints of a LineString, you just need to access its boundary property:

from shapely.geometry import LineString

line = LineString([(0, 0), (1, 1), (2, 2)])
endpoints = line.boundary
print(endpoints)
# MULTIPOINT (0 0, 2 2)
first, last = line.boundary
print(first, last)
# POINT (0 0) POINT (2 2)

Alternatively, you can get the first and the last points from the coords cordinate sequence:

from shapely.geometry import Point
first = Point(line.coords[0])
last = Point(line.coords[-1])
print(first, last)
# POINT (0 0) POINT (2 2)

---

In your specific case, though, as you want to remove the last point of the first line, and the first point of the second line, and only after that get the endpoints, you should construct new LineString objects first using the same coords property:

from shapely.wkt import loads

first_line = loads("LINESTRING (51.2176008 4.4177154, 51.21758 4.4178548, 51.2175729 4.4179023, 51.21745162000732 4.41871738126533)")
second_line = loads("LINESTRING (51.21745162000732 4.41871738126533, 51.2174025 4.4190475, 51.217338 4.4194807, 51.2172511 4.4200562, 51.2172411 4.4201077, 51.2172246 4.4201654, 51.2172067 4.420205, 51.2171806 4.4202355, 51.2171074 4.4202929, 51.2170063 4.4203409, 51.2169564 4.4203641, 51.2168076 4.4204243, 51.2166588 4.4204833, 51.2159018 4.420431, 51.2154117 4.4203843)")
first_line = LineString(first_line.coords[:-1])
second_line = LineString(second_line.coords[1:])
print(first_line.boundary[1], second_line.boundary[0])
# POINT (51.2175729 4.4179023) POINT (51.2174025 4.4190475)

Boleyn

Note, that in the special case where the start and end point of a LineString coincide, the boundary is EMPTY. The approach with accessing line.coords is the only robust solution (in general). – Iatrics

I really like .boundary as a solution, but the syntax has changed in shapely 2.0. See: https://mcmap.net/q/720831/-how-to-get-the-endpoint-of-a-linestring-in-shapely – Entrammel

Answer 2

Note: The syntax changes as of shapely 2.0

Per Georgy's answer, .boundary still works (as of Dec 2022) but is being deprecated and throws a warning.

For multiparts, .geoms is preferred moving forward.

For accessing single points from a coordinate sequence, use numpy.

So, instead of:

first = Point(line.coords[0])
last = Point(line.coords[-1])

use:

first = np.array(line.coords)[0]
last = np.array(line.coords)[-1]

Entrammel

Answer 3

Similar to Georgy's solution, you can get their coords by unpacking the one you want and ignoring the rest using *_.

from shapely.geometry import LineString

linestring1 = LineString([(51.2176008, 4.4177154), (51.21758, 4.4178548),
                         (51.2175729, 4.4179023),
                         (51.21745162000732, 4.41871738126533)])
linestring2 = LineString([(51.21745162000732, 4.41871738126533),
                         (51.2174025, 4.4190475), (51.217338, 4.4194807),
                         (51.2172511, 4.4200562), (51.2172411, 4.4201077),
                         (51.2172246, 4.4201654), (51.2172067, 4.420205),
                         (51.2171806, 4.4202355), (51.2171074, 4.4202929),
                         (51.2170063, 4.4203409), (51.2169564, 4.4203641),
                         (51.2168076, 4.4204243), (51.2166588, 4.4204833),
                         (51.2159018, 4.420431), (51.2154117, 4.4203843)])

*_, last_new_1, last_1 = linestring1.coords
first_2, first_new_2, *_ = linestring2.coords

print(last_new_1)
print(first_new_2)
# (51.2175729, 4.4179023)
# (51.2174025, 4.4190475)

Stirpiculture

Answer 4

Solved with two routines that allows a Linesstring to be split as follows.

1. Function: split_first returns first point, and LineString of points without the first

2. Function: split_last return last point, and LineString of points from first not including the last

Code

from shapely.ops import nearest_points
from shapely.geometry import Point
from shapely.geometry import LineString

def split_first(linestring):
  " returns first point and linestring without first point "
  coords = list(linestring.coords)

  p, *x = coords
  return Point(p), LineString(x)

def split_last(linestring):
  " returns first point and linestring without first point "

  *x, p = list(linestring.coords) = list(linestring.coords)

  return Point(p), LineString(x)

Test

Data

linestring = LineString([(51.2176008,4.4177154), (51.21758,4.4178548), (51.2175729,4.4179023), (51.21745162000732,4.41871738126533)])

First point, and linestring not including the first

p, l = split_first(linestring)
print(p)
print(l)

Out

POINT (51.2176008 4.4177154)
LINESTRING (51.21758 4.4178548, 51.2175729 4.4179023, 51.21745162000732 4.41871738126533)

First last point, and linestring not including the last

p, l = split_last(linestring)
print(p)
print(l)

Out

POINT (51.21745162000732 4.41871738126533)
LINESTRING (51.2176008 4.4177154, 51.21758 4.4178548, 51.2175729 4.4179023)

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