独立移动 numpy 数组的行 [英] Shift rows of a numpy array independently

问题描述

这是在此处提出的问题的扩展(引述如下)

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

我有一个矩阵(准确地说是二维 numpy ndarray):

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

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

我想根据roll独立滚动A的每一行另一个数组中的值:

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

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

也就是说，我想这样做:

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]]

有没有办法有效地做到这一点?也许使用花哨的索引技巧?

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

接受的解决方案是:

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]

我基本上想做同样的事情，除非当索引滚动超过"行尾时，我希望行的另一侧用 NaN 填充，而不是将值移动到以周期性的方式行的前面".

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.

也许以某种方式使用 np.pad ?但我不知道如何让它以不同的数量填充不同的行.

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

推荐答案

Inspired by 独立滚动矩阵行的解决方案，这是一个基于 np.lib.stride_tricks 的矢量化方法.as_strided -

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]

样品运行 -

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]])

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