Create array from mixed floats and arrays Python

Question

Let's say that I have an array x = np.random.normal(size=(300, 300)) and that I wish to create a diagonal matrix of size (2, 2, 300, 300) using x. My first approach was to just do

import numpy as np

x = np.random.normal(size=(300, 300))

array = np.array([
    [x, 0.],
    [0., x]
])

However, when doing so I get an array of size (2, 2). Is there a numpy function for recasting the elements to the same size? Given that all the arrays of the list are the same shape, and that all other elements are floats.

Edit I might add that just defining an empty array and setting the diagonal to x is not a solution.

Edit 2 Here's a sample

import numpy as np


x = np.random.normal(size=(3, 3))

array = np.zeros((2, 2, *x.shape))

array[0, 0] = array[1, 1] = x

print(array)

Yielding

[[[[-1.57346701 -1.00813871 -0.72318135]
   [ 0.11852539  1.144298    1.38860739]
   [ 0.64571669  0.47474236  0.294049  ]]

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


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

  [[-1.57346701 -1.00813871 -0.72318135]
   [ 0.11852539  1.144298    1.38860739]
   [ 0.64571669  0.47474236  0.294049  ]]]]

Use a minimal sample like np.random.normal(size=(3, 3)) and show us the expected output? — Divakar
– Divakar, Commented Dec 4, 2017 at 19:20

Divakar · Accepted Answer · 2017-12-07 11:49:59Z

3

For the case of first two dimension of the output having lengths as 2, the posted solution with - array[0, 0] = array[1, 1] = x would be fast and readable.

I would try to solve for generic lengths.

Approach #1 : Here's one approach with masking -

m = 2 # length along first two axes of o/p
out = np.zeros((m,m) + x.shape)
out[np.eye(m,dtype=bool)] =  x

Approach #2 : Using assignment on first two axes merged view and thus avoiding the creation of any mask and leveraging fast sliced assignment -

out = np.zeros((m,m) + x.shape)
out.reshape((-1,) + x.shape)[::m+1] = x

Generic pattern case

We would leverage np.broadcast_to to solve for such a case -

def resizer(tup):
    shp = x.shape
    out_shp = (len(tup), -1) + shp
    list_arrs = [np.broadcast_to(j, shp) for i in tup for j in i]
    return np.asarray(list_arrs).reshape(out_shp)

Sample run -

In [121]: x
Out[121]: 
array([[55, 58, 75],
       [78, 78, 20],
       [94, 32, 47]])

In [122]: array0 = [
     ...:     [x, 0, 1, 2],
     ...:     [1, 2, 0, x]
     ...: ]

In [123]: resizer(array0)
Out[123]: 
array([[[[55, 58, 75],
         [78, 78, 20],
         [94, 32, 47]],

        [[ 0,  0,  0],
         [ 0,  0,  0],
         [ 0,  0,  0]],

        [[ 1,  1,  1],
         [ 1,  1,  1],
         [ 1,  1,  1]],

        [[ 2,  2,  2],
         [ 2,  2,  2],
         [ 2,  2,  2]]],
        .....

        [[55, 58, 75],
         [78, 78, 20],
         [94, 32, 47]]]])

edited Dec 7, 2017 at 11:49

answered Dec 4, 2017 at 19:25

Divakar

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2 Comments

MaxU - stand with Ukraine Over a year ago

Very simple and elegant!

Tingiskhan Over a year ago

Thank you for this Divakar. As Max says, very simple and elegant (as always)

B. M. · Accepted Answer · 2017-12-04 19:39:52Z

3

You can follow your first idea by:

x = np.random.normal(size=(300, 300))
O = np.zeros_like(x)
r = np.array([[x,O],[O,x]])

In [153]: r.shape
Out[153]: (2, 2, 300, 300)

answered Dec 4, 2017 at 19:39

B. M.

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