你可以利用
numpy.linalg
例程的
矩阵堆叠功能来提高一些速度。这在
numpy.linalg.lstsq
中尚不起作用,但
numpy.linalg.svd可以,因此您可以自己实现lstsq:
import numpy as np
def stacked_lstsq(L, b, rcond=1e-10):
"""
Solve L x = b, via SVD least squares cutting of small singular values
L is an array of shape (..., M, N) and b of shape (..., M).
Returns x of shape (..., N)
"""
u, s, v = np.linalg.svd(L, full_matrices=False)
s_max = s.max(axis=-1, keepdims=True)
s_min = rcond*s_max
inv_s = np.zeros_like(s)
inv_s[s >= s_min] = 1/s[s>=s_min]
x = np.einsum('...ji,...j->...i', v,
inv_s * np.einsum('...ji,...j->...i', u, b.conj()))
return np.conj(x, x)
def slow_lstsq(L, b):
return np.array([np.linalg.lstsq(L[k], b[k])[0]
for k in range(L.shape[0])])
def test_it():
b = np.random.rand(1234, 3)
L = np.random.rand(1234, 3, 6)
x = stacked_lstsq(L, b)
x2 = slow_lstsq(L, b)
print(x.shape, x2.shape)
diff = abs(x - x2).max()
print("difference: ", diff)
assert diff < 1e-13
test_it()
一些时间表明,对于该问题的规模,堆叠版本大约快了6倍。是否值得麻烦取决于问题本身。