Numpy array sequential indexing array of arbitrary dimension

Question

Is it possible to sequentially index points in a numpy array when the number of dimensions n is arbitrary?

Example of desired behavior when `n=4` :

>>> A = np.mgrid[[slice(0,11,1)]*4]
>>> A[XXX_0] # The "first" point in A
array([0,0,0,0])
>>> A[XXX_1] # The "second" point in A
array([0,0,0,1])
>>> A[XXX_11]
array([0,0,1,0])
>>> A[XXX_14640] # The "last" point in A
array([10,10,10,10])

What should replace the above XXX in order to get the above output?

I do not want to convert A to a 2D array, (e.g., using A = A.T.reshape(-1,n)) because this removes the grid structure (produced by mgrid) that is necessary for my application.

MB-F · Accepted Answer · 2018-01-23 15:53:56Z

1

You can use the following variation of A = A.T.reshape(-1,n):

B = A.reshape(n, -1).T

A is not overwritten and its grid structure remains preserved.
No data is copied - B is a view into A (only works because we reshape before .T)

Now you can use A to access the data in a grid structure and B to access the same data in a flat structure.

Example:

A = np.mgrid[[slice(0,11,1)]*4]
B = A.reshape(4, -1).T

# modify first point
print(B[0])  # [0 0 0 0]
B[0] = [1, 3, 3, 7]
print(A[:, 0, 0, 0, 0])  # [1 3 3 7]

answered Jan 23, 2018 at 15:53

MB-F

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Comments

sascha · Accepted Answer · 2018-01-23 15:57:50Z

0

A bit of a potential workaround for some cases (your example is a bit broad):

Code:

import numpy as np

A = np.mgrid[[slice(0,3,1)]*2]
print(A)

access_indices = np.arange(A.size)                       # this one looks like your 0, 1, 2
flat_indices = np.unravel_index(access_indices, A.shape)

A[flat_indices] -= 1
print(A)

Output:

[[[0 0 0]
  [1 1 1]
  [2 2 2]]

 [[0 1 2]
  [0 1 2]
  [0 1 2]]]
[[[-1 -1 -1]
  [ 0  0  0]
  [ 1  1  1]]

 [[-1  0  1]
  [-1  0  1]
  [-1  0  1]]]

But compared to view-based approaches, the only advantage is probably: we don't need to be careful about view vs. copy.

answered Jan 23, 2018 at 15:57

sascha

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Example of desired behavior when `n=4` :