Matrix to Vector with python/numpy

Question

Numpy ravel works well if I need to create a vector by reading by rows or by columns. However, I would like to transform a matrix to a 1d array, by using a method that is often used in image processing. This is an example with initial matrix A and final result B:

A = np.array([[ 0,  1,  2,  3],
              [ 4,  5,  6,  7],
              [ 8,  9, 10, 11],
              [12, 13, 14, 15]])

B = np.array([[ 0, 1, 4, 8, 5, 2, 3, 6, 9, 12, 13, 10, 7, 11, 14, 15])

Is there an existing function already that could help me with that? If not, can you give me some hints on how to solve this problem? PS. the matrix A is NxN.

So it's a zigzag diagonal transversal? Is there a technical name for that? What's the purpose? — hpaulj
– hpaulj, Commented Sep 11, 2016 at 21:18

Warren Weckesser · Accepted Answer · 2016-09-11 23:55:17Z

I've been using numpy for several years, and I've never seen such a function.

Here's one way you could do it (not necessarily the most efficient):

In [47]: a
Out[47]: 
array([[ 0,  1,  2,  3],
       [ 4,  5,  6,  7],
       [ 8,  9, 10, 11],
       [12, 13, 14, 15]])

In [48]: np.concatenate([np.diagonal(a[::-1,:], k)[::(2*(k % 2)-1)] for k in range(1-a.shape[0], a.shape[0])])
Out[48]: array([ 0,  1,  4,  8,  5,  2,  3,  6,  9, 12, 13, 10,  7, 11, 14, 15])

Breaking down the one-liner into separate steps:

a[::-1, :] reverses the rows:

In [59]: a[::-1, :]
Out[59]: 
array([[12, 13, 14, 15],
       [ 8,  9, 10, 11],
       [ 4,  5,  6,  7],
       [ 0,  1,  2,  3]])

(This could also be written a[::-1] or np.flipud(a).)

np.diagonal(a, k) extracts the kth diagonal, where k=0 is the main diagonal. So, for example,

In [65]: np.diagonal(a[::-1, :], -3)
Out[65]: array([0])

In [66]: np.diagonal(a[::-1, :], -2)
Out[66]: array([4, 1])

In [67]: np.diagonal(a[::-1, :], 0)
Out[67]: array([12,  9,  6,  3])

In [68]: np.diagonal(a[::-1, :], 2)
Out[68]: array([14, 11])

In the list comprehension, k gives the diagonal to be extracted. We want to reverse the elements in every other diagonal. The expression 2*(k % 2) - 1 gives the values 1, -1, 1, ... as k varies from -3 to 3. Indexing with [::1] leaves the order of the array being indexed unchanged, and indexing with [::-1] reverses the order of the array. So np.diagonal(a[::-1, :], k)[::(2*(k % 2)-1)] gives the kth diagonal, but with every other diagonal reversed:

In [71]: [np.diagonal(a[::-1,:], k)[::(2*(k % 2)-1)] for k in range(1-a.shape[0], a.shape[0])]
Out[71]: 
[array([0]),
 array([1, 4]),
 array([8, 5, 2]),
 array([ 3,  6,  9, 12]),
 array([13, 10,  7]),
 array([11, 14]),
 array([15])]

np.concatenate() puts them all into a single array:

In [72]: np.concatenate([np.diagonal(a[::-1,:], k)[::(2*(k % 2)-1)] for k in range(1-a.shape[0], a.shape[0])])
Out[72]: array([ 0,  1,  4,  8,  5,  2,  3,  6,  9, 12, 13, 10,  7, 11, 14, 15])

hpaulj · Accepted Answer · 2016-09-11 21:46:28Z

2

I found discussion of zigzag scan for MATLAB, but not much for numpy. One project appears to use a hardcoded indexing array for 8x8 blocks

https://github.com/lot9s/lfv-compression/blob/master/scripts/our_mpeg/zigzag.py

ZIG = np.array([[0,  1,  5,  6,  14, 15, 27, 28],
               [2,  4,  7,  13, 16, 26, 29, 42],
               [3,  8,  12, 17, 25, 30, 41, 43],
               [9,  11, 18, 24, 31, 40, 44,53],
               [10, 19, 23, 32, 39, 45, 52,54],
               [20, 22, 33, 38, 46, 51, 55,60],
               [21, 34, 37, 47, 50, 56, 59,61],
               [35, 36, 48, 49, 57, 58, 62,63]])

Apparently it's used jpeg and mpeg compression.

answered Sep 11, 2016 at 21:46

hpaulj

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