Skip every nth index of numpy array

Question

In order to do K-fold validation I would like to use slice a numpy array such that a view of the original array is made but with every nth element removed.

For example:

[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

If n = 4 then the result would be

[1, 2, 4, 5, 6, 8, 9]

Note: the numpy requirement is due to this being used for a machine learning assignment where the dependencies are fixed.

Answer 1

Approach #1 with modulus

a[np.mod(np.arange(a.size),4)!=0]

Sample run -

In [255]: a
Out[255]: array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])

In [256]: a[np.mod(np.arange(a.size),4)!=0]
Out[256]: array([1, 2, 3, 5, 6, 7, 9])

---

Approach #2 with masking : Requirement as a view

Considering the views requirement, if the idea is to save on memory, we could store the equivalent boolean array that would occupy 8 times less memory on Linux system. Thus, such a mask based approach would be like so -

# Create mask
mask = np.ones(a.size, dtype=bool)
mask[::4] = 0

Here's the memory requirement stat -

In [311]: mask.itemsize
Out[311]: 1

In [312]: a.itemsize
Out[312]: 8

Then, we could use boolean-indexing as a view -

In [313]: a
Out[313]: array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])

In [314]: a[mask] = 10

In [315]: a
Out[315]: array([ 0, 10, 10, 10,  4, 10, 10, 10,  8, 10])

---

Approach #3 with NumPy array strides : Requirement as a view

You can use np.lib.stride_tricks.as_strided to create such a view given the length of the input array is a multiple of n. If it's not a multiple, it would still work, but won't be a safe practice, as we would be going beyond the memory allocated for input array. Please note that the view thus created would be 2D.

Thus, an implementaion to get such a view would be -

def skipped_view(a, n):
    s = a.strides[0]
    strided = np.lib.stride_tricks.as_strided
    return strided(a,shape=((a.size+n-1)//n,n),strides=(n*s,s))[:,1:]

Sample run -

In [50]: a = np.array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]) # Input array

In [51]: a_out = skipped_view(a, 4)

In [52]: a_out
Out[52]: 
array([[ 1,  2,  3],
       [ 5,  6,  7],
       [ 9, 10, 11]])

In [53]: a_out[:] = 100 # Let's prove output is a view indeed

In [54]: a
Out[54]: array([  0, 100, 100, 100,   4, 100, 100, 100,   8, 100, 100, 100])

Answer 2

numpy.delete :

In [18]: arr = np.array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])

In [19]: arr = np.delete(arr, np.arange(0, arr.size, 4))

In [20]: arr
Out[20]: array([1, 2, 3, 5, 6, 7, 9])