给定尺寸的值范围的组合

我使用以下代码创建数组的索引列表。但是，我希望索引按 Fortran 顺序运行，即内部循环是更快的变化循环。有没有办法在 python 中实现这一点。目前，我得到的输出是 C 顺序的。

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

Output:

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

但是，我需要以下形式的输出：

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

A. 两参数解（两列输出）

您可以使用以下命令生成这些索引np.indices然后转置和重塑就可以完成工作 -

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

示例输出 -

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

我们可以使用翻转进一步优化它points with np.indices然后使用np.column_stack创建最终的2列数组。让我们花时间并对照已经提议的方案进行验证。下面列出这两种方法 -

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()))

时间安排 -

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，这个更适合较大的形状。

B. 通用解决方案（通用列输出）

我们将使其通用，以便我们可以使用给定的尽可能多的参数，如下所示 -

def get_combinations(params, order='right'):
    # params : tuple of input scalars that denotes sizes
    # The order arg is used for the LSB position. So, with order='right', the
    # rightmost column is the least significant, hence it will change the most
    # when going through the rows. For order='left', the leftmost column
    # would change the most.
    all_indices = np.indices(params)
    if order=='right':
        return np.moveaxis(all_indices,0,-1).reshape(-1,len(params))
    elif order=='left':
        return all_indices.T.reshape(-1,len(params))
    else:
        raise Exception('Wrong side value!')

示例案例运行 -

In [189]: get_combinations((2,3), order='left')
Out[189]: 
array([[0, 0],
       [1, 0],
       [0, 1],
       [1, 1],
       [0, 2],
       [1, 2]])

In [191]: get_combinations((2,3,2), order='right')
Out[191]: 
array([[0, 0, 0],
       [0, 0, 1],
       [0, 1, 0],
       [0, 1, 1],
       [0, 2, 0],
       [0, 2, 1],
       [1, 0, 0],
       [1, 0, 1],
       [1, 1, 0],
       [1, 1, 1],
       [1, 2, 0],
       [1, 2, 1]])