Is there a way to construct 3D array from 2D array in python using numpy?

Question

I've got a matrix that values are positive integers and want to find the bit representation unpacked into a tensor using numpy

e.g

[[1 , 2],
 [3 , 4]]

to

[[0 , 0],
 [0 , 1]], 

[[0 , 1],
 [1 , 0]],

[[1 , 0],
 [1 , 0]]

Actually I want to map a 2D array that its entries are decimal to binary tensor. In this case tensor is 3 - 2D arrays with binary entries — ASD
– ASD, Commented Sep 14, 2021 at 9:02
depth of tensor is depend on length of max entry , in this case 4 is max entry so 4 equals 100 then depth is 3 — ASD
– ASD, Commented Sep 14, 2021 at 9:04
Something like np.unpackbits(x[None,:], axis=0)[-3:] should do the job. — obchardon
– obchardon, Commented Sep 14, 2021 at 9:42
That's right I looked it but it doesn't support custom unit length. it just works with unit8 — ASD
– ASD, Commented Sep 14, 2021 at 9:44

Michael Szczesny · Accepted Answer · 2021-09-14 11:11:37Z

1

This solution is limited to 2D arrays but works with elements larger than np.uint8 and calculates the necessary bit depth.

import numpy as np

M = np.array([[1,2],
              [3,4]])
bits = int(np.log2(M.max())) + 1

(np.where(
     M.reshape(-1,1) & 2**np.arange(bits)[::-1], 1, 0)
   .reshape(*M.shape, -1)
   .transpose(2,0,1))

Output

array([[[0, 0],
        [0, 1]],

       [[0, 1],
        [1, 0]],

       [[1, 0],
        [1, 0]]])

How this works

Construct a range with powers of 2

2**np.arange(bits)[::-1]

Broadcast this range with logical_and over the input elements

(M.reshape(-1,1) & 2**np.arange(bits)[::-1])

Output

array([[0, 0, 1],
       [0, 2, 0],
       [0, 2, 1],
       [4, 0, 0]])

Convert to 1,0 bool array with np.where

array([[0, 0, 1],   # 1 in binary
       [0, 1, 0],   # 2 in binary
       [0, 1, 1],   # 3 in binary
       [1, 0, 0]])  # 4 in binary

Shape to desired output.

edited Sep 14, 2021 at 11:11

answered Sep 14, 2021 at 10:13

Michael Szczesny

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1 Comment

ASD Over a year ago

Thanks . That's really useful for me :D

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