Numpy数组索引顺序

Question

我正在使用以下代码来创建数组的索引列表.但是,我希望索引以Fortran顺序运行,即内部循环是更快变化的循环.有没有办法在python中实现这一点.目前,我得到的输出是C顺序.

np.transpose(np.nonzero(np.ones([32,30])))

输出:

array([[ 0,  0],
       [ 0,  1],
       [ 0,  2],
       ..., 
       [31, 27],
       [31, 28],
       [31, 29]])

但是,我需要以下形式的ouptut:

array([[ 0,  0],
       [ 1,  0],
       [ 2,  0],
       ..., 
       [29, 29],
       [30, 29],
       [31, 29]])

Answer 1

您可以使用转换生成这些索引,np.indices然后转换并重塑该作业 -

np.indices((32,30)).T.reshape(-1,2)

样品输出 -

In [36]: np.indices((32,30)).T.reshape(-1,2)
Out[36]: 
array([[ 0,  0],
       [ 1,  0],
       [ 2,  0],
       ..., 
       [29, 29],
       [30, 29],
       [31, 29]])

运行时测试 -

In [74]: points = [32,30]

# @218's soln
In [75]: %timeit np.transpose(np.nonzero(np.ones(points[::-1])))[:,::-1]
100000 loops, best of 3: 18.6 µs per loop

In [76]: %timeit np.indices((points)).T.reshape(-1,2)
100000 loops, best of 3: 16.1 µs per loop

In [77]: points = [320,300]

# @218's soln
In [78]: %timeit np.transpose(np.nonzero(np.ones(points[::-1])))[:,::-1]
100 loops, best of 3: 2.14 ms per loop

In [79]: %timeit np.indices((points)).T.reshape(-1,2)
1000 loops, best of 3: 1.26 ms per loop

进一步提升性能

我们可以通过翻转进一步优化它points用np.indices,然后使用np.column_stack以创建最终2列阵列.让我们来反对已经提出的问题进行验证.列出以下两种方法 -

def app1(points):
    return np.indices((points)).T.reshape(-1,2)

def app2(points):
    R,C = np.indices((points[::-1]))
    return np.column_stack((C.ravel(), R.ravel()))

计时 -

In [146]: points = [32,30]

In [147]: np.allclose(app1(points), app2(points))
Out[147]: True

In [148]: %timeit app1(points)
100000 loops, best of 3: 14.8 µs per loop

In [149]: %timeit app2(points)
100000 loops, best of 3: 17.4 µs per loop

In [150]: points = [320,300]

In [151]: %timeit app1(points)
1000 loops, best of 3: 1.1 ms per loop

In [152]: %timeit app2(points)
1000 loops, best of 3: 822 µs per loop

所以,这个更大的形状更好.