X = np.einsum('ik,jk->ij', np.conj(x), x)
相當於
X = np.empty((x.shape[0], x.shape[0]), dtype='complex128')
for i in range(x.shape[0]):
for j in range(x.shape[0]):
X[i, j] = np.vdot(x[i], x[j])
np.einsum
需要的乘積之和。下標'ik,jk->ij'
告訴np.einsum
第二個參數 np.conj(x)
是下標ik
的數組,第三個參數x
具有 下標jk
。因此,產品np.conj(x)[i,k]*x[j,k]
是針對所有的 i
,j
,k
計算的。總和取自重複的下標k
,並且由於 還剩下i
和j
,它們將成爲結果數組的下標。
例如,
import numpy as np
N, M = 10, 20
a = np.random.random((N,M))
b = np.random.random((N,M))
x = a + b*1j
def orig(x):
X = np.empty((x.shape[0], x.shape[0]), dtype='complex128')
for i in range(x.shape[0]):
for j in range(x.shape[0]):
X[i, j] = np.vdot(x[i], x[j])
return X
def alt(x):
return np.einsum('ik,jk->ij', np.conj(x), x)
assert np.allclose(orig(x), alt(x))
In [307]: %timeit orig(x)
10000 loops, best of 3: 143 µs per loop
In [308]: %timeit alt(x)
100000 loops, best of 3: 8.63 µs per loop