In MATLAB there is an easy way to define multidimensional arrays e.g.
A(:,:,1) = [1,2,3; 4,5,6];
A(:,:,2) = [7,8,9; 10,11,12];
>> A
A(:,:,1) =
1 2 3
4 5 6
A(:,:,2) =
7 8 9
10 11 12
where the first two indices are respectively, for the rows and columns of the ith matrix (or page, see picture below) stored in A;
Does anybody know how can I define the same structure in python?
with NumPy indexing is similar to MATLAB
import numpy as np
A=np.empty((2,3,3))
A.shape
#(2L, 3L, 3L)
A[0,1,2] # element at index 0,1,2
#0.0
A[0,:,:] # 3x3 slice at index 0
#array([[ 0., 0., 0.],
# [ 0., 0., 0.],
# [ 0., 0., 0.]])
A[1,1,:] # 1-D array of length 3
#array([ 0., 0., 0.]
import statement since you use np and it makes your answer "more canned". +1.
A(x,y,z) while python follows A[z,y,x] when doing maths
matlab uses 1-based indexing like R instead of python's 0-based indexing.
A pure Python way to do this is using a list of lists (or in this case a list of lists of lists). You can initialize it with list comprehension. For instance:
w = 4 #width
h = 3 #height
d = 3 #depth
data = [[[0]*h for _ in range(w)] for _ in range(d)]
Or if you want to fill the tensor with tuples like on the figure:
data = [[[(i+1,j+1,k+1) for k in range(h)] for j in range(w)] for i in range(d)]
This initializes a d x w x h "matrix" filled with zeros.
You can then access the (i,j,k)-th element with:
data[i][j][k]
Nevertheless there are libraries like numpy that have support for vectors, matrices, tensors, etc.
If you're willing to use NumPy then there's plenty of ways. One way would be to initialise with all zeros or, as in your updated example, you could also fill with a range and then reshape.
import numpy as np
a = np.arange(48, dtype=np.int64).reshape((3, 4, 4))
# or
b = np.zeros((3, 4, 4), dtype=np.int64)
