Combining arrays to yield a new collective array

Question

I have three (n,n) arrays that I need to combine in a very specific way, in order to yield n*n new arrays, that have to be combined into one big array.

I essentially need to take one element from each array and create a new (3,3) array, wherein the diagonal is the three elements (the rest is empty) and then combine these new arrays into one.

It's a bit difficult to explain properly. I've attempted to give an example below which hopefully gives an idea of what I'm trying to do.

Example: Given three (2,3) arrays:

a = np.array([[2,5,9], [7,2,4]])
b = np.array([[3,6,2], [1,6,8]])
c = np.array([[8,7,4], [9,3,1]])

create six arrays with the elements from a, b, and c as the diagonals:

T1 = ([[ 2, 0, 0],
       [ 0, 3, 0],
       [ 0, 0, 8]])

T2 = ([[ 5, 0, 0],
       [ 0, 6, 0],
       [ 0, 0, 7]])

T3 = ([[ 9, 0, 0],
       [ 0, 2, 0],
       [ 0, 0, 4]])

T4 = ([[ 7, 0, 0],
       [ 0, 1, 0],
       [ 0, 0, 9]])

T5 = ([[ 2, 0, 0],
       [ 0, 6, 0],
       [ 0, 0, 3])

T6 = ([[ 4, 0, 0],
       [ 0, 8, 0],
       [ 0, 0, 1]])

combine the six arrays to yield

array([[ 2, 0, 0, 5, 0, 0, 9, 0, 0],
       [ 0, 3, 0, 0, 6, 0, 0, 2, 0],
       [ 0, 0, 8, 0, 0, 7, 0, 0, 4],
       [ 7, 0, 0, 2, 0, 0, 4, 0, 0],
       [ 0, 1, 0, 0, 6, 0, 0, 8, 0],
       [ 0, 0, 9, 0, 0, 3, 0, 0, 1]])

as in

array([[ T1, T2, T3],
       [ T4, T5, T6]])

*The six arrays are not needed in themselves as separate arrays, only the final array is needed. I've just chosen this route as it makes it a bit more apparent what the final one consists of.

Answer 1

It can be done with einsum:

ABC = np.array((a,b,c))
i,j,k = ABC.shape
out = np.zeros((i*j,i*k),ABC.dtype)
np.einsum("jiki->ijk",out.reshape(j,i,k,i))[...] = ABC
out
# array([[2, 0, 0, 5, 0, 0, 9, 0, 0],
#        [0, 3, 0, 0, 6, 0, 0, 2, 0],
#        [0, 0, 8, 0, 0, 7, 0, 0, 4],
#        [7, 0, 0, 2, 0, 0, 4, 0, 0],
#        [0, 1, 0, 0, 6, 0, 0, 8, 0],
#        [0, 0, 9, 0, 0, 3, 0, 0, 1]])

Explanation:

What does the reshape do?

                            axis 2 (size k)
                       /-----------------------\
                            axis 3 (size i)
                       /-----\  /-----\  /-----\   
a  s /     a  s /    [[2, 0, 0, 5, 0, 0, 9, 0, 0],
x  i |     x  i |     [0, 3, 0, 0, 6, 0, 0, 2, 0],
i  z |     i  z \     [0, 0, 8, 0, 0, 7, 0, 0, 4],
s  e |     s  e /     [7, 0, 0, 2, 0, 0, 4, 0, 0],
     |          |     [0, 1, 0, 0, 6, 0, 0, 8, 0],
0  j \     1  i \     [0, 0, 9, 0, 0, 3, 0, 0, 1]]

It isolates the 3x3 diagonal matrices into axes 1,3.

What does einsum do here?

It maps the axes of the reshaped out to those of ABC; "jiki->ijk" means that axis 0 ("j") maps to axis 1, axes 1 and 3 ("i") map to axis 0, and axis 2 ("k") maps to axis 2. Mapping two axes to one (as with "i") has the special meaning of taking the diagonal.

einsum creates a writeable view, so all that's left to do is assigning ABC to that.

Note: that we use the same letters i,j,k for the shape and for the einsum spec doesn't syntactically mean anything, it just makes the thing a lot more readable.

Answer 2

I am unsure as to why you would need to do this, but I believe that I have answered your question anyway. The code is roughly commented, and the variable names are slightly odd, however, it does what you wanted it to do and it does it in the way you suggested above. The code is not very efficient or fast, though it could be cleaned up and made much faster. It takes the arrays you want to convert into the larger output array, makes them the diagonals of 6 3x3 arrays, and then inserts them into the required spot in the output array.

# Import numpy
import numpy as np

# Create your arrays
a = np.array([[2,5,9], [7,2,4]])
b = np.array([[3,6,2], [1,6,8]])
c = np.array([[8,7,4], [9,3,1]])

# Make them into a list
abc = []
abc.append(a)
abc.append(b)
abc.append(c)

# Create an array that will contain T1, T2, ...
arrays = []
for i in range(6):
    arr = np.ndarray(shape=(3, 3))

    # Fill the array with zeros
    for x in range(3):
        for y in range(3):
            arr[x][y] = 0

    for j in range(3):
        arr[j][j] = abc[j][0 if i < 3 else 1][i % 3]

    arrays.append(arr)


# Combine the arrays into one, in the way specified


bigarr = np.ndarray(shape=(6, 9))

offsetX = 0
offsetY = 0
arr = 0

# Loop over all of the arrays (T1, T2, etc.)
for arr in range(6):
    for i in range(3):
        for j in range(3):
            bigarr[i + offsetX][j + offsetY] = arrays[arr][i][j]

    # Offset the place the arrays will be inserted
    offsetY += 3
    if offsetY >= 9:
        offsetY = 0
        offsetX += 3

# The final output is bigarr
print(bigarr)

I hope this answers your question, and if not helps you find another answer.

Answer 3

Here is one way to do that with advanced indexing:

import numpy as np

a = np.array([[2, 5, 9], [7, 2, 4]])
b = np.array([[3, 6, 2], [1, 6, 8]])
c = np.array([[8, 7, 4], [9, 3, 1]])
# Put all input arrays together
abc = np.stack([a, b, c])
# Works with any shape and number of arrays
n, r, c = abc.shape
# Row and column index grid
ii, jj = np.ogrid[:r, :c]
# Shift row and column indices over submatrices of result
idx = np.arange(n)[:, np.newaxis, np.newaxis]
row_idx = ii * n + idx
col_idx = jj * n + idx
# Broadcast indices
row_idx, col_idx = np.broadcast_arrays(row_idx, col_idx)
# Make output
out = np.zeros((n * r, n * c), abc.dtype)
out[row_idx, col_idx] = abc
print(out)
# [[2 0 0 5 0 0 9 0 0]
#  [0 3 0 0 6 0 0 2 0]
#  [0 0 8 0 0 7 0 0 4]
#  [7 0 0 2 0 0 4 0 0]
#  [0 1 0 0 6 0 0 8 0]
#  [0 0 9 0 0 3 0 0 1]]

Answer 4

We can combine the 3 arrays with stack (or np.array):

In [65]: a = np.array([[2,5,9], [7,2,4]]) 
    ...: b = np.array([[3,6,2], [1,6,8]]) 
    ...: c = np.array([[8,7,4], [9,3,1]])                                                                    
In [66]: abc = np.stack((a,b,c))                                                                             
In [67]: abc.shape                                                                                           
Out[67]: (3, 2, 3)

One 'column' of abc is one of your diagonals:

In [68]: abc[:,0,0]                                                                                          
Out[68]: array([2, 3, 8])

Make a target array to hold all 6 diagonals:

In [69]: TT = np.zeros((6,3,3),int)     

We can then set one diagonal with:

In [70]: idx=np.arange(3)                                                                                    
In [71]: TT[0,idx,idx] = abc[:,0,0]                                                                          
In [72]: TT                                                                                                  
Out[72]: 
array([[[2, 0, 0],
        [0, 3, 0],
        [0, 0, 8]],
        ...

To set all 6 we need an array that matches this shape:

In [74]: TT[:,idx,idx].shape                                                                                 
Out[74]: (6, 3)

Reshape abc. The result is (3,6). Transpose to make a (6,3):

In [75]: abc.reshape(3,6)                                                                                    
Out[75]: 
array([[2, 5, 9, 7, 2, 4],
       [3, 6, 2, 1, 6, 8],
       [8, 7, 4, 9, 3, 1]])
In [76]: TT[:,idx,idx] = abc.reshape(3,6).T                                                                  
In [77]: TT                                                                                                  
Out[77]: 
array([[[2, 0, 0],
        [0, 3, 0],
        [0, 0, 8]],

       [[5, 0, 0],
        [0, 6, 0],
        [0, 0, 7]],

       [[9, 0, 0],
        [0, 2, 0],
        [0, 0, 4]],

       [[7, 0, 0],
        [0, 1, 0],
        [0, 0, 9]],

       [[2, 0, 0],
        [0, 6, 0],
        [0, 0, 3]],

       [[4, 0, 0],
        [0, 8, 0],
        [0, 0, 1]]])

Rearrange elements with reshapes and transpose:

In [82]: TT.reshape(2,3,3,3).transpose(0,2,1,3).reshape(6,9)                                                 
Out[82]: 
array([[2, 0, 0, 5, 0, 0, 9, 0, 0],
       [0, 3, 0, 0, 6, 0, 0, 2, 0],
       [0, 0, 8, 0, 0, 7, 0, 0, 4],
       [7, 0, 0, 2, 0, 0, 4, 0, 0],
       [0, 1, 0, 0, 6, 0, 0, 8, 0],
       [0, 0, 9, 0, 0, 3, 0, 0, 1]])

I came up that, step by step. You may want to recreate those steps for yourself. I won't take up the space here.

There may be more direct ways of creating this, but I think the steps are instructive.