NumPy Stack multidimensional arrays

Question

I have 3 nD arrays as follows

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


y = [[10, 11, 12],
     [13, 14, 15],
     [16, 17, 18]]

z = [[ 19,  20,  21],
     [ 22,  23,  24],
     [ 25,  26,  27]]

Without using for loop, I am trying to append every 2x2 matrix elements together such that

a1 = [[1,2]
      [4,5]]

a2 = [[10,11], 
      [13,14]]

a3 = [[19,20],
      [22,23]]

should append to

a = [[1,10,19],[2,11,20],[4,13,22],[5,14,23]]

Please note, the NxN matrix will always be N = j - 1 where j is x.shape(i,j)

Similarly for other 2x2 matrices, the arrays are as follows

b = [[2,11,20],[3,12,21],[5,14,23],[6,15,24]]
c = [[4,13,22],[5,14,23],[7,16,25],[8,17,26]]
d = [[5,14,23],[6,15,24],[8,17,26],[9,18,27]]

For large datasets, the for loop impacts runtime so I am trying to see if there is a way using NumPy stacking techniques

It doesn't answer my question. Please reopen it. I am trying to extract 2x2 from 3 multidimensional arrays into 2d array. I only have x,yz. How do I create all the possible 2x2 matrices .i.e a1, a2, a3, b1, b2, b3 and then append elements ? — Pradeep Tummala
– Pradeep Tummala, Commented Jun 12, 2020 at 8:56
Think that should be your starting point to use np.dstack. Looks like you are sliding across. So, maybe use sckit-image's view_as_windows there. — Divakar
– Divakar, Commented Jun 12, 2020 at 9:01
What is the NxN here? Is it always going to be 2x2 or is it going to be NxN for the smaller matrices? — r4bb1t
– r4bb1t, Commented Jun 12, 2020 at 10:07

hpaulj · Accepted Answer · 2020-06-12 16:07:30Z

Your 3 arrays:

In [46]: x=np.arange(1,10).reshape(3,3)                                                                         
In [48]: y=np.arange(10,19).reshape(3,3)                                                                        
In [49]: z=np.arange(19,28).reshape(3,3)

combined into one:

In [50]: xyz=np.stack((x,y,z))                                                                                  
In [51]: xyz                                                                                                    
Out[51]: 
array([[[ 1,  2,  3],
        [ 4,  5,  6],
        [ 7,  8,  9]],

       [[10, 11, 12],
        [13, 14, 15],
        [16, 17, 18]],

       [[19, 20, 21],
        [22, 23, 24],
        [25, 26, 27]]])

Your a1 (2,2) array:

In [55]: xyz[0,:2,:2]                                                                                           
Out[55]: 
array([[1, 2],
       [4, 5]])

and all 3:

In [56]: xyz[:,:2,:2]                                                                                           
Out[56]: 
array([[[ 1,  2],
        [ 4,  5]],

       [[10, 11],
        [13, 14]],

       [[19, 20],
        [22, 23]]])

and rearrange them into the desired (4,3):

In [57]: xyz[:,:2,:2].transpose(1,2,0)                                                                          
Out[57]: 
array([[[ 1, 10, 19],
        [ 2, 11, 20]],

       [[ 4, 13, 22],
        [ 5, 14, 23]]])
In [58]: xyz[:,:2,:2].transpose(1,2,0).reshape(4,3)                                                             
Out[58]: 
array([[ 1, 10, 19],
       [ 2, 11, 20],
       [ 4, 13, 22],
       [ 5, 14, 23]])

similarly for the other windows:

In [59]: xyz[:,1:3,:2].transpose(1,2,0).reshape(4,3)                                                            
Out[59]: 
array([[ 4, 13, 22],
       [ 5, 14, 23],
       [ 7, 16, 25],
       [ 8, 17, 26]])
In [60]: xyz[:,0:2,1:3].transpose(1,2,0).reshape(4,3)                                                           
Out[60]: 
array([[ 2, 11, 20],
       [ 3, 12, 21],
       [ 5, 14, 23],
       [ 6, 15, 24]])
In [61]: xyz[:,1:3,1:3].transpose(1,2,0).reshape(4,3)                                                           
Out[61]: 
array([[ 5, 14, 23],
       [ 6, 15, 24],
       [ 8, 17, 26],
       [ 9, 18, 27]])

We could also the view_as_windows as @Divakar suggested (or as_strided), but conceptually that's trickier.

====

I can skip the transpose if I stack differently:

In [65]: xyz=np.stack((x,y,z), axis=2)                                                                          
In [66]: xyz                                                                                                    
Out[66]: 
array([[[ 1, 10, 19],
        [ 2, 11, 20],
        [ 3, 12, 21]],

       [[ 4, 13, 22],
        [ 5, 14, 23],
        [ 6, 15, 24]],

       [[ 7, 16, 25],
        [ 8, 17, 26],
        [ 9, 18, 27]]])

In [68]: xyz[:2,:2].reshape(4,3)                                                                                
Out[68]: 
array([[ 1, 10, 19],
       [ 2, 11, 20],
       [ 4, 13, 22],
       [ 5, 14, 23]])

===

In [84]: import skimage                                                                                         
In [85]: skimage.util.view_as_windows(xyz,(2,2,3),1).shape                                                      
Out[85]: (2, 2, 1, 2, 2, 3)
In [86]: skimage.util.view_as_windows(xyz,(2,2,3),1).reshape(4,4,3)                                             
Out[86]: 
array([[[ 1, 10, 19],
        [ 2, 11, 20],
        [ 4, 13, 22],
        [ 5, 14, 23]],

       [[ 2, 11, 20],
        [ 3, 12, 21],
        [ 5, 14, 23],
        [ 6, 15, 24]],

       [[ 4, 13, 22],
        [ 5, 14, 23],
        [ 7, 16, 25],
        [ 8, 17, 26]],

       [[ 5, 14, 23],
        [ 6, 15, 24],
        [ 8, 17, 26],
        [ 9, 18, 27]]])

Sairam pinjarla · Accepted Answer · 2020-06-12 14:00:44Z

0


a1 = np.array([[1,2],[4,5]])

a2 = np.array([[10,11],[13,14]])

a3 = np.array([[19,20],[22,23]])

def everything(a1,a2,a3):

    b1 = a1.reshape(-1)
    b2 = a2.reshape(-1)
    b3 = a3.reshape(-1)
    c = np.concatenate((b1, b2, b3))
    b = [[0, 0, 0], [0, 0, 0], [0, 0, 0], [0, 0, 0]]
    def inner(a, i):
        while i < len(a):
            i = i + 1
            return a[i - 1]
    def looping(a, c):
        k = 0
        j = 0
        while j < len(a) - 1:
            i = 0
            while i < len(a):
                b[i][j] = inner(c, k)
                i += 1
                k += 1
            j += 1
    looping(b3, c)
    print(b)
everything(a1,a2,a3)

edited Jun 12, 2020 at 14:00

answered Jun 12, 2020 at 13:50

Sairam pinjarla

12 bronze badges

1 Comment

rizerphe Over a year ago

Welcome to Stack Overflow! While this code may solve the question, including an explanation of how and why this solves the problem would really help to improve the quality of your post, and probably result in more up-votes. Remember that you are answering the question for readers in the future, not just the person asking now. Please edit your answer to add explanations and give an indication of what limitations and assumptions apply.

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