Reshape and concatenate within a numpy array

Question

I wanted to know if it exists an easy way to transform such a matrix :

[[[1, 2, 3, 4], [5, 6, 7, 8]], [[9, 10, 11, 12], [13, 14, 15, 16]]]

into

[[1 ,2 ,5 ,6 ],
 [3 ,4 ,7 ,8 ],
 [9 ,10,13,14],
 [11,12,15,16]]

Which is equivalent to reshape each initial list into 2x2 matrices, and then concatenate them; for instance np.array([1,2,3,4]).reshape((2,2)) gives [[1,2],[3,4]]

np.array([5, 6, 7, 8]).reshape((2,2)) gives [[5,6],[7,8]]

so

np.concatenate((np.array([1,2,3,4]).reshape((2,2)), np.array([5, 6, 7, 8]).reshape((2,2))), axis=1)

will give

array([[1 ,2 ,5 ,6 ],
       [3 ,4 ,7 ,8 ]])

etc...

It is indeed a dummy example since I need to deal with more (and bigger) matrices, I have to find a more straightforward method.

If I understand it correctly, you want to "drop" the second dimension? — willeM_ Van Onsem
– willeM_ Van Onsem, Commented Feb 8, 2020 at 22:10

yatu · Accepted Answer · 2020-02-08 23:20:35Z

1

We can use reshape with swapaxes and a concatenate along the first axis:

np.concatenate(a.reshape(a.shape[0], a.shape[1], 2, -1)
                 .swapaxes(1,2)
                 .reshape(a.shape))

array([[ 1,  2,  5,  6],
       [ 3,  4,  7,  8],
       [ 9, 10, 13, 14],
       [11, 12, 15, 16]])

edited Feb 8, 2020 at 23:20

answered Feb 8, 2020 at 22:18

yatu

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6 Comments

Netchaiev Over a year ago

Wow! That's impressive! Thank you so much! Now I just have to learn and understand this swapaxes method I didn't know :)

Netchaiev Over a year ago

Finaly I couldn't apply it to my problem... I had Q.shape = (10,10,784), and I wanted, in the same way as in my example to find a (280,280) shaped matrix ; I ended up to use

data = np.array([])  for l in Q:      ligne = np.array([])     for s in l:         if ligne.shape == (0,) :             ligne = s.reshape((28,28))         else:             ligne = np.concatenate((ligne, s.reshape((28,28))), axis=1)     if data.shape == (0,) :         data = ligne     else:         data = np.concatenate((data, ligne), axis=0)

Netchaiev Over a year ago

What are, in this case, a,b,c,d,e for np.concatenate(Q.reshape(a, b, c, -1).swapaxes(d,e).reshape(Q.shape)) ?

yatu Over a year ago

My bad I hardcoded the two first axes, try now @Netchaiev

Netchaiev Over a year ago

Still not working (I obtain at the end a 100x784 matrix...), since in my matrix of interest Q, Q.shape[0] and Q.shape[1] will be 10 and, at some point I guess one must to turn the 784 long lists into 28x28 matrices...

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