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
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!
Try using the norm function from
d = norm(np.cross(p2-p1, p1-p3))/norm(p2-p1)
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
which for an array of points
p3 will give you an array of distances from the line.
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)
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.
To find distance to line from point if you have slope and intercept you can use formula from wiki
def distance(point,coef): return abs((coef*point)-point+coef)/math.sqrt((coef*coef)+1)
coef is a tuple with slope and intercept
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 B = coef C = A*x + B*x
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 B = coef C = A*x + B*x shortest_dis = shortest_distance(x, y, A, B, C)
Please let me know if any of this is unclear.
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
- x-intercept p1 = [0, -4/3]
- y-intercept p2 = [2, 0]
- shortest distance from p3 = [5, 6] = 3.328
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
3D distance should use np.dot
for j in range (0,len(pre_points_3_d)): vec1 = list(map(lambda x:x- x,zip(pre_points_3_d[j], points_camera))) vec2 = list(map(lambda x:x- x,zip(pre_points_3_d[j], points_3_d[j]))) vec3 = list(map(lambda x:x- x,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