Coordinates of the closest points of two geometries in Shapely

The GIS term you are describing is linear referencing, and Shapely has these methods.

# Length along line that is closest to the pointprint(line.project(p))# Now combine with interpolated point on linep2 = line.interpolate(line.project(p))print(p2)  # POINT (5 7)

An alternative method is to use nearest_points:

from shapely.ops import nearest_pointsp2 = nearest_points(line, p)[0]print(p2)  # POINT (5 7)

which provides the same answer as the linear referencing technique does, but can determine the nearest pair of points from more complicated geometry inputs, like two polygons.

In case you have a single segment (e.g.: a line, as referring to the title) rather than a list of segments, here is what I did, and with a passing test case. Please consider that some users on this page are looking just for that from looking at the title, coming from a Google search.

Python code:

def sq_shortest_dist_to_point(self, other_point):    dx = self.b.x - self.a.x    dy = self.b.y - self.a.y    dr2 = float(dx ** 2 + dy ** 2)    lerp = ((other_point.x - self.a.x) * dx + (other_point.y - self.a.y) * dy) / dr2    if lerp < 0:        lerp = 0    elif lerp > 1:        lerp = 1    x = lerp * dx + self.a.x    y = lerp * dy + self.a.y    _dx = x - other_point.x    _dy = y - other_point.y    square_dist = _dx ** 2 + _dy ** 2    return square_distdef shortest_dist_to_point(self, other_point):    return math.sqrt(self.sq_shortest_dist_to_point(other_point))

A test case:

def test_distance_to_other_point(self):    # Parametrize test with multiple cases:    segments_and_point_and_answer = [        [Segment(Point(1.0, 1.0), Point(1.0, 3.0)), Point(2.0, 4.0), math.sqrt(2.0)],        [Segment(Point(1.0, 1.0), Point(1.0, 3.0)), Point(2.0, 3.0), 1.0],        [Segment(Point(0.0, 0.0), Point(0.0, 3.0)), Point(1.0, 1.0), 1.0],        [Segment(Point(1.0, 1.0), Point(3.0, 3.0)), Point(2.0, 2.0), 0.0],        [Segment(Point(-1.0, -1.0), Point(3.0, 3.0)), Point(2.0, 2.0), 0.0],        [Segment(Point(1.0, 1.0), Point(1.0, 3.0)), Point(2.0, 3.0), 1.0],        [Segment(Point(1.0, 1.0), Point(1.0, 3.0)), Point(2.0, 4.0), math.sqrt(2.0)],        [Segment(Point(1.0, 1.0), Point(-3.0, -3.0)), Point(-3.0, -4.0), 1],        [Segment(Point(1.0, 1.0), Point(-3.0, -3.0)), Point(-4.0, -3.0), 1],        [Segment(Point(1.0, 1.0), Point(-3.0, -3.0)), Point(1, 2), 1],        [Segment(Point(1.0, 1.0), Point(-3.0, -3.0)), Point(2, 1), 1],        [Segment(Point(1.0, 1.0), Point(-3.0, -3.0)), Point(-3, -1), math.sqrt(2.0)],        [Segment(Point(1.0, 1.0), Point(-3.0, -3.0)), Point(-1, -3), math.sqrt(2.0)],        [Segment(Point(-1.0, -1.0), Point(3.0, 3.0)), Point(3, 1), math.sqrt(2.0)],        [Segment(Point(-1.0, -1.0), Point(3.0, 3.0)), Point(1, 3), math.sqrt(2.0)],        [Segment(Point(1.0, 1.0), Point(3.0, 3.0)), Point(3, 1), math.sqrt(2.0)],        [Segment(Point(1.0, 1.0), Point(3.0, 3.0)), Point(1, 3), math.sqrt(2.0)]    ]    for i, (segment, point, answer) in enumerate(segments_and_point_and_answer):        result = segment.shortest_dist_to_point(point)        self.assertAlmostEqual(result, answer, delta=0.001, msg=str((i, segment, point, answer)))

Note: I assume this function is inside a Segment class.In case your line is infinite, don't limit the lerp from 0 to 1 only, but still at least provide two distinct a and b points.