python - Distance between point and a line (from two points)

26
votes

I'm using Python+Numpy (can maybe also use Scipy) and have three 2D points 

(P1, P2, P3);

I am trying to get the distance from P3 perpendicular to a line drawn between P1 and P2. Let P1=(x1,y1), P2=(x2,y2) and P3=(x3,y3)

In vector notation this would be pretty easy, but I'm fairly new to python/numpy and can't get anythng that works (or even close).

Any tips appreciated, thanks!

8
pythonnumpyvectorscipypoint

8 Answers

42
votes

Try using the norm function from numpy.linalg

d = norm(np.cross(p2-p1, p1-p3))/norm(p2-p1)
12
votes

np.cross returns the z-coordinate of the cross product only for 2D vectors. So the first norm in the accepted answer is not needed, and is actually dangerous if p3 is an array of vectors rather than a single vector. Best just to use

d=np.cross(p2-p1,p3-p1)/norm(p2-p1)

which for an array of points p3 will give you an array of distances from the line.

10
votes

For the above-mentioned answers to work, the points need to be numpy arrays, here's a working example:

import numpy as np
p1=np.array([0,0])
p2=np.array([10,10])
p3=np.array([5,7])
d=np.cross(p2-p1,p3-p1)/np.linalg.norm(p2-p1)
3
votes
abs((x2-x1)*(y1-y0) - (x1-x0)*(y2-y1)) / np.sqrt(np.square(x2-x1) + np.square(y2-y1))

Can be used directly through the formula, just have to plug in the values and boom it will work.

3
votes

To find distance to line from point if you have slope and intercept you can use formula from wiki https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_line Python:

def distance(point,coef):
    return abs((coef[0]*point[0])-point[1]+coef[1])/math.sqrt((coef[0]*coef[0])+1)

coef is a tuple with slope and intercept

1
votes

Based on the accepted answer

Test with below line equation -

Find the perpendicular distance from the point (5, 6) to the line −2x + 3y + 4 = 0

import numpy as np
norm = np.linalg.norm

p1 = np.array([0,-4/3])
p2 = np.array([2, 0])

p3 = np.array([5, 6])
d = np.abs(norm(np.cross(p2-p1, p1-p3)))/norm(p2-p1)
# output d = 3.328201177351375
0
votes

Shortest Distance from Point to a Line

This is the code I got from https://www.geeksforgeeks.org:

import math 

# Function to find distance 
def shortest_distance(x1, y1, a, b, c):    
    d = abs((a * x1 + b * y1 + c)) / (math.sqrt(a * a + b * b)) 
    print("Perpendicular distance is", d)

Now you have to find A, B, C, x, and y.

import numpy as np
closest = []
x = (x ,y)
y = (x, y)
coef = np.polyfit(x, y, 1)
A = coef[0]
B = coef[1]
C = A*x[0] + B*x[1]

Now you can plug in the values:

shortest_dis = shortest_distance(x, y, A, B, C)

The full code may look like this:

import math
import numpy as np

def shortest_distance(x1, y1, a, b, c):    
    d = abs((a * x1 + b * y1 + c)) / (math.sqrt(a * a + b * b)) 
    print("Perpendicular distance is", d)

closest = []
x = (x ,y)
y = (x, y)
coef = np.polyfit(x, y, 1)
A = coef[0]
B = coef[1]
C = A*x[0] + B*x[1]
shortest_dis = shortest_distance(x, y, A, B, C)

Please let me know if any of this is unclear.

-1
votes

3D distance should use np.dot def threeD_corres(points_3_d,pre_points_3_d,points_camera):

  for j in  range (0,len(pre_points_3_d)):
      vec1 = list(map(lambda x:x[0]- x[1],zip(pre_points_3_d[j], points_camera)))
      vec2 = list(map(lambda x:x[0]- x[1],zip(pre_points_3_d[j], points_3_d[j])))
      vec3 =  list(map(lambda x:x[0]- x[1],zip(points_3_d[j], points_camera)))
      distance = np.abs(np.dot(vec1_1,vec2_2))/np.linalg.norm(vec3)

      print("#########distance:\n",distance)
  return  distance