How to select elements from a multidimensional array with python : array[:][i][j]?

Question

In Python, I have a 3 dimensional array A :

A=np.random.random((3,2,3))

print(A)

Out[6]: [[[0.89565135 0.79502322 0.89015957]
  [0.40303084 0.80496231 0.50278239]]

 [[0.70610822 0.61269108 0.00470925]
  [0.29734101 0.67986295 0.34584381]]

 [[0.71822397 0.99326199 0.40949422]
  [0.97733739 0.38916931 0.91475145]]]

I would like to select the first element of each submatrix and to make an array from them [0.89565135,0.70610822,0.71822397] so I tried the following formula : A[:][0][0],A[0][:][0],A[0][0][:] but they all give me the same result, which is not the one I'm expecting...

A[:][0][0]
Out[7]: array([0.89565135, 0.79502322, 0.89015957])

A[0][:][0]
Out[8]: array([0.89565135, 0.79502322, 0.89015957])

A[0][0][:]
Out[9]: array([0.89565135, 0.79502322, 0.89015957])

What formula can I use to get the right array and how come the above formula give the same result ?

Compare A[:,0,0] to A[0,:,0] and A[0,0,:]. Still identical? — Learning is a mess
– Learning is a mess, Commented Dec 6, 2021 at 11:12

Antoine Dubuis · Accepted Answer · 2021-12-06 11:24:14Z

2

You can do as follows to select for all channel, the first element of the matrix:

A[:, 0, 0]

: references all channels, 0 the first row and 0 the first column.

Output:

array([0.89565135,0.70610822,0.71822397])

EDIT:

If your are working with nested list, you can do as follows:

[array[0][0]for array in A]

This shows one of the benefit of numpy array over python list as the selection is much more easier (and faster) using numpy.

edited Dec 6, 2021 at 11:24

answered Dec 6, 2021 at 11:17

Antoine Dubuis

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2 Comments

J.A Over a year ago

But still if I write the array directly not as a numpy array :

A=[[[0.89565135, 0.79502322, 0.89015957], [0.40303084, 0.80496231, 0.50278239]],  [[0.70610822, 0.61269108, 0.00470925], [0.29734101, 0.67986295, 0.34584381]],  [[0.71822397, 0.99326199, 0.40949422], [0.97733739, 0.38916931, 0.91475145]]]

the problem remains. Do you have an idea in that case ?

Antoine Dubuis Over a year ago

I added a solution using nested list. However i would advise to perform this operation with numpy as it is much more efficient.

phoney_badger · Accepted Answer · 2021-12-06 11:24:37Z

1

Your idea for indexing is a bit flawed here, when you're doing A[:][0][0] what you're doing is A[:] would mean basically the whole matrix, out of which you're specifying out the first index using A[:][0] which would be

[[0.89565135 0.79502322 0.89015957]
  [0.40303084 0.80496231 0.50278239]]

out of which you're further specifying the first index A[:][0][0] which will give you the first row

[0.89565135 0.79502322 0.89015957]

Following a similar logic you can see that in your second indexing scheme you're effectively doing the same thing. What you need to do is to do

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

which you can write in short hand

A[:,0,0]

answered Dec 6, 2021 at 11:24

phoney_badger

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