Python 添加多个矩阵而不构建新矩阵

假设我有两个矩阵

B

和

M

，我想执行以下语句：

B += 3*M

我反复执行此指令，因此不希望每次矩阵

3*M

时都构建它（

3

可能会发生变化，只是为了确保我只生成标量矩阵积）。是不是一个numpy函数让这个计算“就位”

更准确地说，我有一个标量列表

as

和一个矩阵列表

Ms

，我想执行这两个的“点积”（因为两个操作数的类型不同，所以不是一个），也就是说：

sum(a*M for a, M in zip(as, Ms))

np.dot

函数除了…

您可以使用-

或-

样本运行-

In [41]: As = [2,5,6]

In [42]: Ms = [np.random.rand(2,3),np.random.rand(2,3),np.random.rand(2,3)]

In [43]: sum(a*M for a, M in zip(As, Ms))
Out[43]: 
array([[  6.79630284,   5.04212877,  10.76217631],
       [  4.91927651,   1.98115548,   6.13705742]])

In [44]: np.tensordot(As,Ms,axes=(0,0))
Out[44]: 
array([[  6.79630284,   5.04212877,  10.76217631],
       [  4.91927651,   1.98115548,   6.13705742]])

In [45]: np.einsum('i,ijk->jk',As,Ms)
Out[45]: 
array([[  6.79630284,   5.04212877,  10.76217631],
       [  4.91927651,   1.98115548,   6.13705742]])

---

另一种方法是利用

因此，您可以从1D和2D阵列创建3D阵列，然后在适当的轴上求和：

>>> Ms = np.random.randn(4, 2, 3)   # 4 arrays of size 2x3
>>> As = np.random.randn(4)
>>> np.sum(As[:, np.newaxis, np.newaxis] * Ms)
array([[-1.40199248, -0.40337845, -0.69986566],
       [ 3.52724279,  0.19547118,  2.1485559 ]])
>>> sum(a*M for a, M in zip(As, Ms))
array([[-1.40199248, -0.40337845, -0.69986566],
       [ 3.52724279,  0.19547118,  2.1485559 ]])

然而，值得注意的是，

np.einsum

和

np.tensordot

通常效率更高：

>>> %timeit np.sum(As[:, np.newaxis, np.newaxis] * Ms, axis=0)
The slowest run took 7.38 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 8.58 µs per loop
>>> %timeit np.einsum('i,ijk->jk', As, Ms)
The slowest run took 19.16 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 2.44 µs per loop

对于更多的人来说也是如此：

>>> Ms = np.random.randn(100, 200, 300)
>>> As = np.random.randn(100)
>>> %timeit np.einsum('i,ijk->jk', As, Ms)
100 loops, best of 3: 5.03 ms per loop
>>> %timeit np.sum(As[:, np.newaxis, np.newaxis] * Ms, axis=0)
100 loops, best of 3: 14.8 ms per loop
>>> %timeit np.tensordot(As,Ms,axes=(0,0))
100 loops, best of 3: 2.79 ms per loop

因此

np.tensordot

在这种情况下效果最好

使用

np.sum

和广播的唯一好理由是使代码更具可读性（当矩阵较小时会有所帮助）

>>> %timeit np.sum(As[:, np.newaxis, np.newaxis] * Ms, axis=0)
The slowest run took 7.38 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 8.58 µs per loop
>>> %timeit np.einsum('i,ijk->jk', As, Ms)
The slowest run took 19.16 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 2.44 µs per loop

>>> Ms = np.random.randn(100, 200, 300)
>>> As = np.random.randn(100)
>>> %timeit np.einsum('i,ijk->jk', As, Ms)
100 loops, best of 3: 5.03 ms per loop
>>> %timeit np.sum(As[:, np.newaxis, np.newaxis] * Ms, axis=0)
100 loops, best of 3: 14.8 ms per loop
>>> %timeit np.tensordot(As,Ms,axes=(0,0))
100 loops, best of 3: 2.79 ms per loop