How to make a 2d numpy array a 3d array?

Question

I have a 2d array with shape (x, y) which I want to convert to a 3d array with shape (x, y, 1). Is there a nice Pythonic way to do this?

Answer 1

hope this funtion helps u to convert 2D array to 3D array.

Args:
  x: 2darray, (n_time, n_in)
  agg_num: int, number of frames to concatenate. 
  hop: int, number of hop frames. 

Returns:
  3darray, (n_blocks, agg_num, n_in)


def d_2d_to_3d(x, agg_num, hop):

    # Pad to at least one block. 
    len_x, n_in = x.shape
    if (len_x < agg_num): #not in get_matrix_data
        x = np.concatenate((x, np.zeros((agg_num - len_x, n_in))))

    # main 2d to 3d. 
    len_x = len(x)
    i1 = 0
    x3d = []
    while (i1 + agg_num <= len_x):
        x3d.append(x[i1 : i1 + agg_num])
        i1 += hop

    return np.array(x3d)

Answer 2

numpy.reshape(array, array.shape + (1,))

Answer 3

You can do this with reshape

For example, you have an array A of shape 35 x 750 (two dimensions), you can change the shape to 35 x 25 x 30 (three dimensions) with A.reshape(35, 25, 30)

More in the documentation here

Answer 4

import numpy as np

# create a 2D array
a = np.array([[1,2,3], [4,5,6], [1,2,3], [4,5,6],[1,2,3], [4,5,6],[1,2,3], [4,5,6]])

print(a.shape) 
# shape of a = (8,3)

b = np.reshape(a, (8, 3, -1)) 
# changing the shape, -1 means any number which is suitable

print(b.shape) 
# size of b = (8,3,1)

Answer 5

import numpy as np

a= np.eye(3)
print a.shape
b = a.reshape(3,3,1)
print b.shape

Answer 6

In addition to the other answers, you can also use slicing with numpy.newaxis:

>>> from numpy import zeros, newaxis
>>> a = zeros((6, 8))
>>> a.shape
(6, 8)
>>> b = a[:, :, newaxis]
>>> b.shape
(6, 8, 1)

Or even this (which will work with an arbitrary number of dimensions):

>>> b = a[..., newaxis]
>>> b.shape
(6, 8, 1)