Accessing NumPy array elements not in a given index list

The way you have your data, the simplest approach is to use np.delete:

sub_array = np.delete(array, index, axis=2)

Alternatively, the logical operators you were trying to use can be applied with boolean arrays as @DSM suggests:

mask = np.ones(a.shape[2], dtype=bool)
mask[index] = False
sub_array = array[:,:, mask]

(I wouldn't call your array array but I followed the names in your question)

The question is answered but I propose a benchmark of the three methods here.

Fastest solution is boolean mask (with small and larger index array size)

mask = np.ones(arr.size, dtype=bool)
mask[indexes] = False
result = arr[mask]

It is 2000 times faster than the list comprehension and marginaly faster than np.delete

Code to reproduce

Three proposed solutions: list comprehension (sol1), boolean mask (sol2) or np.delete (sol3)

d = 100000
a = np.random.rand(d)
idx = np.random.randint(d, size = 10)


# list comprehension
def sol1(arr, indexes):
    return arr[[i for i in range(arr.size) if i not in indexes]]
sol1(a, idx)
# Out[30]: array([0.13044518, 0.68564961, 0.03033223, ..., 0.03796257, 0.40137137, 0.45403929])

# boolean mask
def sol2(arr, indexes):
    mask = np.ones(arr.size, dtype=bool)
    mask[indexes] = False
    return arr[mask]
sol2(a, idx)
# Out[32]: array([0.13044518, 0.68564961, 0.03033223, ..., 0.03796257, 0.40137137, 0.45403929])

# np.delete
def sol3(arr, indexes):
    return np.delete(arr, indexes)
sol3(a, idx)
# Out[36]: array([0.13044518, 0.68564961, 0.03033223, ..., 0.03796257, 0.40137137, 0.45403929])

Results

%timeit sol1(a, idx)
384 ms ± 2.75 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%timeit sol2(a, idx)
154 µs ± 15.7 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit sol3(a, idx)
194 µs ± 18.8 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)


idx = np.random.randint(d, size = 1000)
%timeit sol1(a, idx)
386 ms ± 7.75 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%timeit sol2(a, idx)
171 µs ± 11.3 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit sol3(a, idx)
205 µs ± 10.5 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)