Slice every item except every nth

Question

In tensorflow it is possible to select every nth item with the slicing notation [::n].

But how to do the opposite? I want to select every item except every nth.

For example:

a = [1, 2, 3, 4, 5, 6, 7, 8]

a[2::3] would result in [3, 6]

Now I would like to have the opposite: [1, 2, 4, 5, 7, 8]

The array above is just an example. A solution should work for bigger matrices of the dimension [batch, width, height, channels] in tensorflow. The selection is only done on the channels. Also my matrix contains real values that are non unique. I will also not be able to reshape it further down than to two dimensions ( [batch, channels] )

Answer 1

One option is to create a boolean index by testing the a range index:

import numpy as np
start, step = 2, 3
a[np.arange(len(a)) % step != start]
# array([1, 2, 4, 5, 7, 8])

You can achieve this similarly in tensorflow using tf.boolean_mask:

import tensorflow as tf

a = tf.constant([1, 2, 3, 4, 5, 6, 7, 8])

start, step = 2, 3
mask = ~tf.equal(tf.range(a.shape[-1]) % step, start)

tf.boolean_mask(a, mask).eval()
# array([1, 2, 4, 5, 7, 8], dtype=int32)

If a is ND tensor, you can specify the axis with boolean_mask; with 4D tensor [batch, width, height, channels] for instance, to select by the fourth axis, i.e the channels, you can set axis=3:

mask = ~tf.equal(tf.range(a.shape[-1]) % step, start)
tf.boolean_mask(a, mask, axis=3)

Answer 2

You can use the np.delete() method:

>>> np.delete(a, a[1::3])
array([1, 2, 4, 5, 7, 8])

Remembering that this operation does not modify the original array:

Return a new array with sub-arrays along an axis deleted. For a one dimensional array, this returns those entries not returned by arr[obj].