Creating 3D Array in Python

Question

I am trying to form a 3D array in Python by populating it with 2D arrays. N is a number than varies depending on the file being read. The matrix is forming as 3D but only appears to have 1 'layer' to it when I am expecting it to have N layers. It appears that the N number of 'layers' is not being passed into the formed array.

import numpy as np

#'rot' is a 3D matrix of shape (N,3,3)
a=np.array(rot[:,0,0])
b=np.array(rot[:,0,1])
c=np.array(rot[:,0,2])
d=np.array(rot[:,1,0])
e=np.array(rot[:,1,1])
f=np.array(rot[:,1,2])
g=np.array(rot[:,2,0])
h=np.array(rot[:,2,1])
i=np.array(rot[:,2,2])  

print(a.shape)
#(N,)

#Forming 3D array
arr=np.array([[[a,b,1],
               [d,e,1],
               [g,h,i]]])

print(arr.shape)
#(1,3,3)

Does this answer your question? [Efficiently Creating A Pandas DataFrame From A Numpy 3d array](https://stackoverflow.com/questions/36235180/efficiently-creating-a-pandas-dataframe-from-a-numpy-3d-array) — Zaraki Kenpachi, Feb 26 '20 at 14:57
What is the shape that you are expecting? The `1` entries are likely causing a problem since the array now has vectors of length `N` and integers. Try `np.array([1]*N)` to get a vector of 1s with the same shape as the other elements. — Chris Mueller, Feb 26 '20 at 15:18
Did you look at `arr`? Or `arr.dtype`. Don't just look at `shape`, especially when the shape is unexpected. — hpaulj, Feb 26 '20 at 18:01
The initial size 1 dimension is produced by the outer layer of `[]`. — hpaulj, Feb 26 '20 at 19:41

score -1 · Answer 1 · answered Feb 26 '20 at 15:23

You don't say exactly what shape you are expecting. The code below will return a 3D array of shape (3, 3, N)

ones_vec = np.array([1] * N)

arr = np.array([[a, b, ones_vec], 
                [d, e, ones_vec], 
                [g, h, h]])
print(arr.shape)
# (3, 3, N)

hpaulj · Answer 2 · 2020-02-26T20:46:48.197

Let's simplify the case, and look at the actual results, not just the shape.

In [327]: a=np.arange(4)

Make an array from just this array (or others like it)

In [328]: np.array([[[a,a],[a,a]]])                                                            
Out[328]: 
array([[[[0, 1, 2, 3],
         [0, 1, 2, 3]],

        [[0, 1, 2, 3],
         [0, 1, 2, 3]]]])
In [329]: _.shape                                                                              
Out[329]: (1, 2, 2, 4)

Note the shape and dtype (int like a). But when we add a scalar 1 to the mix:

In [330]: np.array([[[a,a,1],[a,a,1]]])                                                        
Out[330]: 
array([[[array([0, 1, 2, 3]), array([0, 1, 2, 3]), 1],
        [array([0, 1, 2, 3]), array([0, 1, 2, 3]), 1]]], dtype=object)
In [331]: _.shape                                                                              
Out[331]: (1, 2, 3)

The dtype has changed. It is now a mix of arrays and the scalar element.

Replacing the scalar with an array that matches a:

In [332]: ones=np.ones_like(a)                                                                 
In [333]: ones                                                                                 
Out[333]: array([1, 1, 1, 1])
In [334]: np.array([[[a,a,ones],[a,a,ones]]])                                                  
Out[334]: 
array([[[[0, 1, 2, 3],
         [0, 1, 2, 3],
         [1, 1, 1, 1]],

        [[0, 1, 2, 3],
         [0, 1, 2, 3],
         [1, 1, 1, 1]]]])
In [335]: _.shape           
Out[335]: (1, 2, 3, 4)

and without the outer layer of []

In [356]: np.array([[a,a,ones],[a,a,ones]])                                                    
Out[356]: 
array([[[0, 1, 2, 3],
        [0, 1, 2, 3],
        [1, 1, 1, 1]],

       [[0, 1, 2, 3],
        [0, 1, 2, 3],
        [1, 1, 1, 1]]])
In [357]: _.shape                                                                              
Out[357]: (2, 3, 4)

If you want the N (here 4) dimension you can transpose

  arr.transpose(2,0,1)

Creating 3D Array in Python

2 Answers2