Shift rows of a numpy array independently

Question

This is an extension of the question posed here (quoted below)

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?

The accepted solution was:

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]

I would basically like to do the same thing, except when an index gets rolled "past" the end of the row, I would like the other side of the row to be padded with a NaN, rather than the value move to the "front" of the row in a periodic fashion.

Maybe using np.pad somehow? But I can't figure out how to get that to pad different rows by different amounts.

Answer 1

Inspired by Roll rows of a matrix independently's solution, here's a vectorized one based on np.lib.stride_tricks.as_strided -

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]

Sample run -

In [76]: a
Out[76]: 
array([[4, 0, 0],
       [1, 2, 3],
       [0, 0, 5]])

In [77]: r
Out[77]: array([ 2,  0, -1])

In [78]: strided_indexing_roll(a, r)
Out[78]: 
array([[nan, nan,  4.],
       [ 1.,  2.,  3.],
       [ 0.,  5., nan]])