Numpy:计算2d阵列每行对角线的最快方法

Question

给定一个2d Numpy数组,我希望能够以最快的方式计算每一行的对角线,我现在正在使用列表理解,但我想知道它是否可以以某种方式进行矢量化？

例如,使用以下M数组:

M = np.random.rand(5, 3)


[[ 0.25891593  0.07299478  0.36586996]
 [ 0.30851087  0.37131459  0.16274825]
 [ 0.71061831  0.67718718  0.09562581]
 [ 0.71588836  0.76772047  0.15476079]
 [ 0.92985142  0.22263399  0.88027331]]

我想计算以下数组:

np.array([np.diag(row) for row in M])

array([[[ 0.25891593,  0.        ,  0.        ],
        [ 0.        ,  0.07299478,  0.        ],
        [ 0.        ,  0.        ,  0.36586996]],

       [[ 0.30851087,  0.        ,  0.        ],
        [ 0.        ,  0.37131459,  0.        ],
        [ 0.        ,  0.        ,  0.16274825]],

       [[ 0.71061831,  0.        ,  0.        ],
        [ 0.        ,  0.67718718,  0.        ],
        [ 0.        ,  0.        ,  0.09562581]],

       [[ 0.71588836,  0.        ,  0.        ],
        [ 0.        ,  0.76772047,  0.        ],
        [ 0.        ,  0.        ,  0.15476079]],

       [[ 0.92985142,  0.        ,  0.        ],
        [ 0.        ,  0.22263399,  0.        ],
        [ 0.        ,  0.        ,  0.88027331]]])

Answer 1

这是使用np.eye(3)(3x3身份数组)的元素乘法和稍微重新形状的一种方式M:

>>> M = np.random.rand(5, 3)
>>> np.eye(3) * M[:,np.newaxis,:]
array([[[ 0.42527357,  0.        ,  0.        ],
        [ 0.        ,  0.17557419,  0.        ],
        [ 0.        ,  0.        ,  0.61920924]],

       [[ 0.04991268,  0.        ,  0.        ],
        [ 0.        ,  0.74000307,  0.        ],
        [ 0.        ,  0.        ,  0.34541354]],

       [[ 0.71464307,  0.        ,  0.        ],
        [ 0.        ,  0.11878955,  0.        ],
        [ 0.        ,  0.        ,  0.65411844]],

       [[ 0.01699954,  0.        ,  0.        ],
        [ 0.        ,  0.39927673,  0.        ],
        [ 0.        ,  0.        ,  0.14378892]],

       [[ 0.5209439 ,  0.        ,  0.        ],
        [ 0.        ,  0.34520876,  0.        ],
        [ 0.        ,  0.        ,  0.53862677]]])

(通过"重新塑造M",我的意思是使各行M沿z轴面向外而不是横跨y轴,从而M形成形状(5, 1, 3).)

Answer 2

尽管@ajcr得到了很好的答案,但使用花式索引可以实现更快的替代方案(在NumPy 1.9.0中测试):

import numpy as np

def sol0(M):
    return np.eye(M.shape[1]) * M[:,np.newaxis,:]

def sol1(M):
    b = np.zeros((M.shape[0], M.shape[1], M.shape[1]))
    diag = np.arange(M.shape[1])
    b[:, diag, diag] = M
    return b

时间显示这大约快4倍:

M = np.random.random((1000, 3))
%timeit sol0(M)
#10000 loops, best of 3: 111 µs per loop
%timeit sol1(M)
#10000 loops, best of 3: 23.8 µs per loop