Basically given some vector v, I want to get another random vector w with some cosine similarity between v and w. Is there any way we can get this in python?
Example: for simplicity I will have 2D vector of v [3,-4]. I want to get random vector w with cosine similarity of 60% or plus 0.6. This should generate vector w with values [0.875, 3] or any other vector with same cosine similarity. So I hope this is clear enough.
Given the vector v and cosine similarity costheta (a scalar between -1 and 1), compute w as in the function rand_cos_sim(v, costheta):
import numpy as np
def rand_cos_sim(v, costheta):
# Form the unit vector parallel to v:
u = v / np.linalg.norm(v)
# Pick a random vector:
r = np.random.multivariate_normal(np.zeros_like(v), np.eye(len(v)))
# Form a vector perpendicular to v:
uperp = r - r.dot(u)*u
# Make it a unit vector:
uperp = uperp / np.linalg.norm(uperp)
# w is the linear combination of u and uperp with coefficients costheta
# and sin(theta) = sqrt(1 - costheta**2), respectively:
w = costheta*u + np.sqrt(1 - costheta**2)*uperp
return w
For example,
In [17]: v = np.array([3, -4])
In [18]: w = rand_cos_sim(v, 0.6)
In [19]: w
Out[19]: array([-0.28, -0.96])
Verify the cosine similarity:
In [20]: v.dot(w)/(np.linalg.norm(v)*np.linalg.norm(w))
Out[20]: 0.6000000000000015
In [21]: w = rand_cos_sim(v, 0.6)
In [22]: w
Out[22]: array([1., 0.])
In [23]: v.dot(w)/(np.linalg.norm(v)*np.linalg.norm(w))
Out[23]: 0.6
The return value always has magnitude 1, so in the above example, there are only two possible random vectors, [1, 0] and [-0.28, -0.96].
Another example, this one in 3-d:
In [24]: v = np.array([3, -4, 6])
In [25]: w = rand_cos_sim(v, -0.75)
In [26]: w
Out[26]: array([ 0.3194265 , 0.46814873, -0.82389531])
In [27]: v.dot(w)/(np.linalg.norm(v)*np.linalg.norm(w))
Out[27]: -0.75
In [28]: w = rand_cos_sim(v, -0.75)
In [29]: w
Out[29]: array([-0.48830063, 0.85783797, -0.16023891])
In [30]: v.dot(w)/(np.linalg.norm(v)*np.linalg.norm(w))
Out[30]: -0.75