Find coordinate of the closest point on polygon in Shapely

Question

Say I have the following Polygon and Point:

>>> poly = Polygon([(0, 0), (2, 8), (14, 10), (6, 1)])
>>> point = Point(12, 4)

I can calculate the point's distance to the polygon...

>>> dist = point.distance(poly)
>>> print(dist)
2.49136439561

...but I would like to know the coordinate of the point on the polygon border where that shortest distance measures to.

My initial approach is to buffer the point by its distance to the polygon, and find the point at which that circle is tangent to the polygon:

>>> buff = point.buffer(dist) 

However, I'm not sure how to calculate that point. The two polygon's don't intersect so list(poly.intersection(buff)) will not give me that point.

Am I on the right track with this? Is there a more straightforward method?

Answer 1

While the answer of eguaio does the job, there is a more natural way to get the closest point using shapely.ops.nearest_points function:

from shapely.geometry import Point, Polygon
from shapely.ops import nearest_points

poly = Polygon([(0, 0), (2, 8), (14, 10), (6, 1)])
point = Point(12, 4)
# The points are returned in the same order as the input geometries:
p1, p2 = nearest_points(poly, point)
print(p1.wkt)
# POINT (10.13793103448276 5.655172413793103)

The result is the same as in the other answer:

from shapely.geometry import LinearRing
pol_ext = LinearRing(poly.exterior.coords)
d = pol_ext.project(point)
p = pol_ext.interpolate(d)
print(p.wkt)
# POINT (10.13793103448276 5.655172413793103)
print(p.equals(p1))
# True