问题描述
我有一个TF-IDF矩阵,形状为(149,1001)。想要的是计算所有列的最后一列的余弦相似度
这就是我所做的
from numpy import dot
from numpy.linalg import norm
for i in range(mat.shape[1]-1):
cos_sim = dot(mat[:,i],mat[:,-1])/(norm(mat[:,i])*norm(mat[:,-1]))
cos_sim
但是此循环使其变慢。那么,有什么有效的方法吗?我只想用numpy做
解决方法
将2D
矢量化的matrix-multiplication
这是使用NumPy在2D数据上使用矩阵乘法的人-
p1 = mat[:,-1].dot(mat[:,:-1])
p2 = norm(mat[:,:-1],axis=0)*norm(mat[:,-1])
out1 = p1/p2
说明::p1
是dot(mat[:,i],mat[:,-1])
循环的矢量化等效项。 p2
属于(norm(mat[:,i])*norm(mat[:,-1]))
。
运行示例以进行验证-
In [57]: np.random.seed(0)
...: mat = np.random.rand(149,1001)
In [58]: out = np.empty(mat.shape[1]-1)
...: for i in range(mat.shape[1]-1):
...: out[i] = dot(mat[:,-1])/(norm(mat[:,-1]))
In [59]: p1 = mat[:,:-1])
...: p2 = norm(mat[:,-1])
...: out1 = p1/p2
In [60]: np.allclose(out,out1)
Out[60]: True
时间-
In [61]: %%timeit
...: out = np.empty(mat.shape[1]-1)
...: for i in range(mat.shape[1]-1):
...: out[i] = dot(mat[:,-1]))
18.5 ms ± 977 µs per loop (mean ± std. dev. of 7 runs,100 loops each)
In [62]: %%timeit
...: p1 = mat[:,-1])
...: out1 = p1/p2
939 µs ± 29.2 µs per loop (mean ± std. dev. of 7 runs,1000 loops each)
# @yatu's soln
In [89]: a = mat
In [90]: %timeit cosine_similarity(a[None,:,-1],a.T[:-1])
2.47 ms ± 461 µs per loop (mean ± std. dev. of 7 runs,100 loops each)
使用norm
在einsum
上进行进一步优化
或者,我们可以用p2
来计算np.einsum
。
因此,norm(mat[:,axis=0)
可以替换为:
np.sqrt(np.einsum('ij,ij->j',:-1]))
因此,给我们修改后的p2
:
p2 = np.sqrt(np.einsum('ij,:-1]))*norm(mat[:,-1])
与以前相同的设置上的计时-
In [82]: %%timeit
...: p1 = mat[:,:-1])
...: p2 = np.sqrt(np.einsum('ij,-1])
...: out1 = p1/p2
607 µs ± 132 µs per loop (mean ± std. dev. of 7 runs,1000 loops each)
30x+
加快了循环速度!
有一个sklearn函数来计算向量cosine_similarity
之间的余弦相似度。这是一个带有示例数组的用例:
a = np.random.randint(0,10,(5,5))
print(a)
array([[5,2,4,1],[4,8,4],[9,7,9,7],6,1,3],[1,5,0]])
from sklearn.metrics.pairwise import cosine_similarity
cosine_similarity(a[None,a.T[:-1])
# array([[0.94022805,0.91705665,0.75592895,0.79921221,1. ]])
a[None,-1]
是a
中的最后一列,其形状经过了重塑,以使两个矩阵具有相等的形状Mat.shape[1]
,这是函数的要求:
a[None,-1]
# array([[1,3,0]])
通过换位,结果将是cosine_similarity
和所有其他列。
检查问题的解决方案:
from numpy import dot
from numpy.linalg import norm
cos_sim = []
for i in range(a.shape[1]-1):
cos_sim.append(dot(a[:,a[:,-1])/(norm(a[:,i])*norm(a[:,-1])))
np.allclose(cos_sim,cosine_similarity(a[None,a.T[:-1]))
# True