tuple as index in multidimensional array together with slicing

Question

I have a 3-dimensional array A , and I need to access A[:,1,1] through the tuple x=[1,1] . Something like this:

x = [1,1]
A[:,*x]

However, doing that I get a syntax error. I would love to be able to access the elements of A[:,1,1] using the variable x , how can I do that?

Thank you!

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Second question:

How to do the same but instead of slicing : do it with a boolean array. For example if t is an array of booleans, obtain A[t, *x]

Answer 1

You can do the following:

import numpy as np

A = np.arange(12).reshape((2, 3, 2))
print(A)

x = [1, 1]
print(A[(slice(None), *x)])

You can use slice(None) instead of : to build a tuple of slices. The tuple environment allows for value unpacking with the * operator.

Output:

[[[ 0  1]
  [ 2  3]
  [ 4  5]]

 [[ 6  7]
  [ 8  9]
  [10 11]]]

[3 9]

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To verify it matches:

import numpy as np

A = np.arange(12).reshape((2, 3, 2))
x = [1, 1]
s = (slice(None), *x)
print(np.allclose(A[s], A[:, 1, 1]))  # True

*This is a modification of answers found here: Slicing a numpy array along a dynamically specified axis

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Edit to reflect edit on question and comment:

To clarify, you can unpack any iterable you like in the tuple environment. The * operator functions normally in within the tuple. Order your elements however you like. Mix in different iterables, types, slice(None) , how ever you want to build your slices, as long as you end up with a valid sequence of values, it will behave as expected.

import numpy as np

A = np.arange(12).reshape((2, 3, 2))
t = [True, False]
x = [1, 1]
print(np.allclose(A[(*t, *x)], A[True, False, 1, 1]))  # True

You can also add full lists as well in the tuple:

print(np.allclose(A[(t, *x)], A[[True, False], 1, 1]))  # True

Answer 2

you can use slice(None) instead of : , So

y = tuple([slice[None]] + x)
A[y]

is what you need.