基本上给定了向量v,我想得到另一个随机向量w,在v和w之间有一些余弦相似性。有什么办法可以在python中获得它?
示例:为简单起见,我将具有v [3,-4]的2D向量。我想获得余弦相似度为60%或加0.6的随机向量w。这将生成具有值[0.875,3]的向量w或具有相同余弦相似度的任何其他向量。所以我希望这足够清楚。
答案 0 :(得分:3)
给定向量v
和余弦相似度costheta
(-1和1之间的标量),像函数w
中一样计算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
例如,
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])
验证余弦相似度:
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
返回值始终为1,因此在上面的示例中,只有两个可能的随机向量[1,0]和[-0.28,-0.96]。
另一个示例,这是3-d中的一个:
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
答案 1 :(得分:-1)
SciPy余弦距离:https://docs.scipy.org/doc/scipy-0.14.0/reference/generated/scipy.spatial.distance.cosine.html
from scipy.spatial.distance import cosine
v = [3, -4]
w = [0.875, 3]
cosine(v, w)
就向后工作而言,您可以使用点积产品自行完成。