Shift rows in array independently

Question

I want to shift each row by its row-number and be relative to my desired output shape. An example:

array([[0, 1, 2],        array([[0, 1, 2],                array([[0, 1, 2, 0, 0],
       [1, 2, 3],    ->            [1, 2, 3],      ->            [0, 1, 2, 3, 0],
       [2, 3, 4]])                    [2, 3, 4]])                [0, 0, 2, 3, 4])

The array to furthest left is my input and the array to the furthest right is my desired output. This can be generalized to bigger arrays, for example a 10x10 array.

Is there a nice way of doing this?

What I have is:

A = np.array([[1, 2, 3],
              [2, 3, 4],
              [3, 4, 5]], dtype=np.float32)

out = np.zeros((A.shape[0], A.shape[1]*2-1))

out[[np.r_[:3]], [np.r_[:3] + np.r_[:3][:,None]]] = A

Javier A · Answer 1 · 2021-04-30T13:54:00.223

2

Here is a solution for square matrices of size n.

np.concatenate((A,np.zeros((n,n))),axis=1).flatten()[0:-n].reshape([n,2*n-1])

edited Apr 30 '21 at 13:54

answered Apr 30 '21 at 13:44

Javier A

539
2
12

1

Thanks! This seems faster than using advanced indexing. I also noticed that your solution would be a bit faster by using ```numpy.ravel``` instead of ```np.flatten```. – Kevin Apr 30 '21 at 13:58

Pierre D · Accepted Answer · 2021-04-30T14:11:39.287

Here is a fast way:

def f(a):
    r, _ = a.shape
    return np.c_[a, np.zeros((r, r), dtype=a.dtype)].ravel()[:-r].reshape(r, -1)

Example:

a = np.arange(8).reshape(4, -1)
>>> f(a)
array([[0, 1, 0, 0, 0],
       [0, 2, 3, 0, 0],
       [0, 0, 4, 5, 0],
       [0, 0, 0, 6, 7]])

Timing

a = np.random.randint(100, size=(1000, 1000))
%timeit f(a)
# 2.08 ms ± 14.5 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

Shift rows in array independently

2 Answers2

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