独立地移动numpy数组的行

Question

这是提出的问题here的延伸（引自下面）

I have a matrix (2d numpy ndarray, to be precise):

A = np.array([[4, 0, 0],
              [1, 2, 3],
              [0, 0, 5]])

And I want to roll each row of A independently, according to roll values in another array:

r = np.array([2, 0, -1])

That is, I want to do this:

print np.array([np.roll(row, x) for row,x in zip(A, r)])

[[0 0 4]
 [1 2 3]
 [0 5 0]]

Is there a way to do this efficiently? Perhaps using fancy indexing tricks?

公认的解决方案是：

rows, column_indices = np.ogrid[:A.shape[0], :A.shape[1]]

# Use always a negative shift, so that column_indices are valid.
# (could also use module operation)
r[r < 0] += A.shape[1]
column_indices = column_indices - r[:,np.newaxis]

result = A[rows, column_indices]

基本上我想做同样的事情，除了当一个索引“超过”行的末尾时，我希望行的另一端用NaN填充，而不是值以周期性的方式移动到行的“前面”。在

也许用np.pad的方法？但我不知道如何让它以不同的数量填充不同的行。在

Answer 1

受Roll rows of a matrix independently's solution的启发，这里有一个基于^{}-

from skimage.util.shape import view_as_windows as viewW

def strided_indexing_roll(a, r):
    # Concatenate with sliced to cover all rolls
    p = np.full((a.shape[0],a.shape[1]-1),np.nan)
    a_ext = np.concatenate((p,a,p),axis=1)

    # Get sliding windows; use advanced-indexing to select appropriate ones
    n = a.shape[1]
    return viewW(a_ext,(1,n))[np.arange(len(r)), -r + (n-1),0]

样本运行-

^{pr2}$

Answer 2

我可以把它和线性索引一起破解…它得到了正确的结果，但在大型数组上执行得相当慢。在

A = np.array([[4, 0, 0],
              [1, 2, 3],
              [0, 0, 5]]).astype(float)

r = np.array([2, 0, -1])

rows, column_indices = np.ogrid[:A.shape[0], :A.shape[1]]

# Use always a negative shift, so that column_indices are valid.
# (could also use module operation)
r_old = r.copy()
r[r < 0] += A.shape[1]
column_indices = column_indices - r[:,np.newaxis]

result = A[rows, column_indices]

# replace with NaNs
row_length = result.shape[-1]

pad_inds = []
for ind,i in np.enumerate(r_old):
    if i > 0:
        inds2pad = [np.ravel_multi_index((ind,) + (j,),result.shape) for j in range(i)]
        pad_inds.extend(inds2pad)
    if i < 0:
        inds2pad = [np.ravel_multi_index((ind,) + (j,),result.shape) for j in range(row_length+i,row_length)]
        pad_inds.extend(inds2pad)
result.ravel()[pad_inds] = nan

给出预期结果：

^{pr2}$