Calculate angle between three points python

I have studied the dot product from vector analysis in my school. Now that formula, I will use for finding the angle between three points. We have use multiple dimentional data like 1D, 2D, 3D and higher dimensions not only 2D. But i explained with 2D data points.

The dot product may be defined algebraically or geometrically. The geometric definition is based on the notions of angle and distance [magnitude of vectors]. The equivalence of these two definitions relies on having a Cartesian coordinate system for Euclidean space.

Geometric definition:

geometric object that possesses both a magnitude and a direction. A vector can be pictured as an arrow. Its magnitude is its length, and its direction is the direction that the arrow points to. The magnitude of a vector a is denoted by ||a||. The dot product of two Euclidean vectors a and b is defined by. where θ is the angle between a and b.

Explanation:

Coding[Python]:

Using math library:

Using numpy library:

I have write down a code to calculate angle between three points using their 3D coordinates.

import  numpy as np

a = np.array[[32.49, -39.96,-3.86]]

b = np.array[[31.39, -39.28, -4.66]]

c = np.array[[31.14, -38.09,-4.49]]

f = a-b # normalization of vectors
e = b-c # normalization of vectors

angle = dot[f, e] # calculates dot product 
print degrees[cos[angle]]  # calculated angle in radians to degree 

output of the code:

degree 33.4118214995

but when i used one of the software to calculate the same it gives output bit different 120 degree. please help

reference i have used to write the program:

[How to calculate bond angle in protein db file?]

Mr_and_Mrs_D

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asked Feb 3, 2016 at 11:58

3

Your original code is pretty close. Adomas.m's answer is not very idiomatic numpy:

import numpy as np

a = np.array[[32.49, -39.96,-3.86]]
b = np.array[[31.39, -39.28, -4.66]]
c = np.array[[31.14, -38.09,-4.49]]

ba = a - b
bc = c - b

cosine_angle = np.dot[ba, bc] / [np.linalg.norm[ba] * np.linalg.norm[bc]]
angle = np.arccos[cosine_angle]

print np.degrees[angle]

answered Feb 3, 2016 at 13:50

EricEric

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2

I guess numpy is quite enough:

    from numpy import *
    from numpy.linalg import norm
    a = array[[32.49, -39.96,-3.86]]
    b = array[[31.39, -39.28, -4.66]]
    c = array[[31.14, -38.09,-4.49]]
    f = b-a 
    e = b-c 
    abVec = norm[f]
    bcVec = norm[e]
    abNorm = f / abVec;
    bcNorm = e / bcVec;
    res = abNorm[0] * bcNorm[0] + abNorm[1] * bcNorm[1] + abNorm[2] * bcNorm[2];
    angle = arccos[res]*180.0/ pi
    print angle

also the res can be calculated with dot:

    res = abNorm[0] * bcNorm[0] + abNorm[1] * bcNorm[1] + abNorm[2] * bcNorm[2];
    res = dot[abNorm, bcNorm]

answered Feb 3, 2016 at 12:18

adomas.madomas.m

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4

For 2D, you can use this method using the math library.

 import math
 
def getAngle[a, b, c]:
    ang = math.degrees[math.atan2[c[1]-b[1], c[0]-b[0]] - math.atan2[a[1]-b[1], a[0]-b[0]]]
    return ang + 360 if ang < 0 else ang
 
print[getAngle[[5, 0], [0, 0], [0, 5]]]

Credits: //manivannan-ai.medium.com/find-the-angle-between-three-points-from-2d-using-python-348c513e2cd

answered Jan 25, 2021 at 18:23

0

In case you have a big list of [x,y,z] coordinates, this works:

import numpy
def compute_angle_between_3d_points[a,b,c]:
    ba = a - b
    bc = c - b

    cosine_numerator = np.sum[ba*bc, axis=1]
    cosine_denominator_1 = np.linalg.norm[ba, axis=1]
    cosine_denominator_2 = np.linalg.norm[bc, axis=1]
    cosine_angle = cosine_numerator / [cosine_denominator_1 * cosine_denominator_2]
    angles = np.arccos[cosine_angle]
    degree_angles = np.rad2deg[angles]

    return degree_angles

Above, a,b,c are presumed to be of shape [N_Points, 3]. Something in TensorFlow or Torch would surely be faster, but there you go.

answered Aug 19 at 22:46

legellegel

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