Python: Appending dimensions to a bidimensional array

Question

Suppose you have an array (m, m) and want to make it (n, n). For example, transforming a 2x2 matrix to a 6x6. So:

[[ 1.  2.]
 [ 3.  4.]]

To:

[[ 1.  2.  0.  0.  0.  0.]
 [ 3.  4.  0.  0.  0.  0.]
 [ 0.  0.  0.  0.  0.  0.]
 [ 0.  0.  0.  0.  0.  0.]
 [ 0.  0.  0.  0.  0.  0.]
 [ 0.  0.  0.  0.  0.  0.]]

This is what I'm doing:

def array_append(old_array, new_shape):
    old_shape = old_array.shape
    dif = np.array(new_shape) - np.array(old_array.shape)
    rows = []
    for i in xrange(dif[0]):
        rows.append(np.zeros((old_array.shape[0])).tolist())
    new_array = np.append(old_array, rows, axis=0)
    columns = []
    for i in xrange(len(new_array)):
        columns.append(np.zeros(dif[1]).tolist())
    return np.append(new_array, columns, axis=1)

Example use:

test1 = np.ones((2,2))
test2 = np.zeros((6,6))
print array_append(test1, test2.shape)

Output:

[[ 1.  1.  0.  0.  0.  0.]
 [ 1.  1.  0.  0.  0.  0.]
 [ 0.  0.  0.  0.  0.  0.]
 [ 0.  0.  0.  0.  0.  0.]
 [ 0.  0.  0.  0.  0.  0.]
 [ 0.  0.  0.  0.  0.  0.]]

Based on this answer. But that's a lot of code for an (imho) simple operation. Is there a more concise/pythonic way to do it?

@pnodnda: your approach is way too complicated. Simply allocate new array and copy old to appropriate location. It's just that simple, as demonstrated on Benjamins (modified) and my answer. Btw the word append is commonly associated with dynamic data structures, which numpy.array is not. Thanks — eat
– eat, Commented Feb 1, 2011 at 21:27

Benjamin · Accepted Answer · 2011-02-01 21:16:33Z

3

Why not use array = numpy.zeros((6,6)), see the numpy docs...

EDIT, woops, question has been edited... I guess you are trying to put ones in a section of an array filled with zeros? Then:

array = numpy.zeros((6,6))
array[0:2,0:2] = 1

If the small matrix does not all have the value of 1:

array[ystart:yend,xstart:xend] = smallermatrix

edited Feb 1, 2011 at 21:16

answered Feb 1, 2011 at 21:05

Benjamin

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5 Comments

pnodbnda Over a year ago

I only added input/output to the code I wrote. You should read the code before replying. And I'm not trying to put ones in an array filled with zeros. I'm taking a small "matrix" and increasing its dimension by padding/filling it with zeros.

Benjamin Over a year ago

In that case, create a blank matrix of the correct size with numpy.zeros and insert your smaller matrix as I show above.

Mia Clarke Over a year ago

@pnodbnda You should maybe entertain the possibility that @Benjamin answered as he did, not to spite you or to be lazy, but because your question is lacking in clarity. The headline is about the only thing that is clear about it, and obviously it doesn't even reflect that well on the content of the question. StackOverflow is a community built on manners and professionalism, please do try to practice that in future remarks on people's efforts.

pnodbnda Over a year ago

@Benjamin Your code assumes the whole matrix has the same value. It doesn't.

pnodbnda Over a year ago

@Banang Yes, I apologize. But hurrying to answer a question without reading it and then trying to pretend you've actually answered anything when you haven't isn't exactly the right attitude either. I'll see if I can edit the OP to make it clearer.

eat · Accepted Answer · 2011-02-01 21:15:49Z

1

That would be then:

# test1= np.ones((2, 2))
test1= np.random.randn((2, 2))
test2= np.zeros((6, 6))
test2[0: 2, 0: 2]= test1

answered Feb 1, 2011 at 21:15

eat

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