大集合上的 Numpy 余弦相似度差异

Question

我需要在大矩阵上使用 Scikit-learn sklearn.metric.pairwise.cosine_similarity。 对于一些优化，我只需要计算矩阵的一些行，所以我尝试了不同的方法。

我发现在某些情况下结果因向量的大小而异，我在这个测试用例（大向量、转置和估计余弦）上看到了这种奇怪的行为： p>

from sklearn.metrics.pairwise import cosine_similarity
from scipy import spatial
import numpy as np
from scipy.sparse import csc_matrix

size=200
a=np.array([[1,0,1,0]]*size)
sparse_a=csc_matrix(a.T)
#standard cosine similarity between the whole transposed matrix, take only the first row
res1=cosine_similarity(a.T,a.T)[0]
#take the row obtained by the multiplication of the first row of the transposed matrix with transposed matrix itself (optimized for the first row calculus only)
res2=cosine_similarity([a.T[0]],a.T)[0]
#sparse matrix implementation with the transposed, which should be faster
res3=cosine_similarity(sparse_a,sparse_a)[0]
print("res1: ",res1)
print("res2: ",res2)
print("res3: ",res3)
print("res1 vs res2: ",res1==res2)
print("res1 vs res3: ",res1==res3)
print("res2 vs res3: ", res2==res3)

如果 "size" 设置为 200 我得到了这个结果，没关系：

res1:  [ 1.  0.  1.  0.]
res2:  [ 1.  0.  1.  0.]
res3:  [ 1.  0.  1.  0.]
res1 vs res2:  [ True  True  True  True]
res1 vs res3:  [ True  True  True  True]
res2 vs res3:  [ True  True  True  True]

但如果“size”设置为2000或更多，就会发生一些奇怪的事情：

res1:  [ 1.  0.  1.  0.]
res2:  [ 1.  0.  1.  0.]
res3:  [ 1.  0.  1.  0.]
res1 vs res2:  [False  True False  True]
res1 vs res3:  [False  True False  True]
res2 vs res3:  [ True  True  True  True]

有人知道我错过了什么吗？

提前致谢

Answer 1

为了比较numpy.array，您必须使用np.isclose 而不是相等运算符。试试：

from sklearn.metrics.pairwise import cosine_similarity
from scipy import spatial
import numpy as np
from scipy.sparse import csc_matrix

size=2000
a=np.array([[1,0,1,0]]*size)
sparse_a=csc_matrix(a.T)
#standard cosine similarity between the whole transposed matrix, take only the first row
res1=cosine_similarity(a.T,a.T)[0]
#take the row obtained by the multiplication of the first row of the transposed matrix with transposed matrix itself (optimized for the first     row calculus only)
res2=cosine_similarity([a.T[0]],a.T)[0]
#sparse matrix implementation with the transposed, which should befaster
res3=cosine_similarity(sparse_a,sparse_a)[0]
print("res1: ",res1)
print("res2: ",res2)
print("res3: ",res3)
print("res1 vs res2: ", np.isclose(res1, res2))
print("res1 vs res3: ", np.isclose(res1, res3))
print("res2 vs res3: ", np.isclose(res2, res2))

结果是：

res1:  [ 1.  0.  1.  0.]
res2:  [ 1.  0.  1.  0.]
res3:  [ 1.  0.  1.  0.]
res1 vs res2:  [ True  True  True  True]
res1 vs res3:  [ True  True  True  True]
res2 vs res3:  [ True  True  True  True]

正如预期的那样。