Numpy repeating a row or column

Question

Suppose we have the matrix A:

A = [1,2,3
     4,5,6
     7,8,9]

I want to know if there is a way to obtain:

B = [1,2,3
     4,5,6
     7,8,9
     7,8,9]

As well as:

B = [1,2,3,3
     4,5,6,6
     7,8,9,9]

This is because the function I want to implement is the following:

U(i,j) = min(A(i+1,j)^2, A(i,j)^2)
V(i,j) = min(A(i,j+1)^2, A(i,j)^2)

And the numpy.minimum seems to need two arrays with equal shapes.

My idea is the following:

np.minimum(np.square(A[1:]), np.square(A[:]))

but it will fail.

xnx · Accepted Answer · 2018-12-06 11:10:15Z

2

For your particular example you could use numpy.hstack and numpy.vstack:

In [11]: np.vstack((A, A[-1]))
Out[11]: 
array([[1, 2, 3],
       [4, 5, 6],
       [7, 8, 9],
       [7, 8, 9]])

In [12]: np.hstack((A, A[:, [-1]]))
Out[12]: 
array([[1, 2, 3, 3],
       [4, 5, 6, 6],
       [7, 8, 9, 9]])

An alternative to the last one is np.hstack((A, np.atleast_2d(A[:,-1]).T)) or np.vstack((A.T, A.T[-1])).T): you can't hstack a (3,) array to a (3,3) one without putting the elements in the rows of a (3,1) array.

edited Dec 6, 2018 at 11:10

answered May 13, 2015 at 18:32

xnx

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2 Comments

Uttam Pal Over a year ago

What about np.hstack((A, A[:, [2]]))

xnx Over a year ago

@UttamPal Good suggestion! I've edited in this approach in a slightly more general form as the prefered one.

tom10 · Accepted Answer · 2015-05-13 19:10:18Z

2

A good answer to your literal question is provided by @xnx, but I wonder whether what you actually need is something else.

This type of question comes up a lot in comparisons, and the usual solution is to take only the valid entries, rather than using a misaligned comparison. That is, something like this is common:

import numpy as np

A = np.arange(9).reshape((3,3))

U = np.minimum(A[1:,:]**2, A[:-1,:]**2)
V = np.minimum(A[:,1:]**2, A[:,:-1]**2)

print U
# [[ 0  1  4]
#  [ 9 16 25]]

print V
# [[ 0  1]
#  [ 9 16]
#  [36 49]]

I suspect that you're probably thinking, "that's a hassle, now U and V have different shapes than A, which is not what I want". But, to this I'd say, "yes, it is a hassle, but it's better to deal with the problem up front and directly than hide it within an invalid row of an array."

A standard example and typical use case of this approach would be numpy.diff, where, "the shape of the output is the same as a except along axis where the dimension is smaller by n."

edited May 13, 2015 at 19:10

answered May 13, 2015 at 18:40

tom10

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2 Comments

FacundoGFlores Over a year ago

Thank you for your advice. Let me see if I catch it: Do you mean I have to do something like np.minimum(A[1:], A[:-1]) and repeat the last row after that?

tom10 Over a year ago

Yes, exactly. I wrote the wrong expression. I'll fix it now.

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