Coordinates of the closest points of two geometries in Shapely

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_dist

def 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.

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

# Length along line that is closest to the point
print(line.project(p))

# Now combine with interpolated point on line
p2 = line.interpolate(line.project(p))
print(p2)  # POINT (5 7)

---

An alternative method is to use nearest_points:

from shapely.ops import nearest_points
p2 = 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.