You can use numpy's slicing, simply start:stop:step.

>>> xs
array([1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4])
>>> xs[1::4]
array([2, 2, 2])

This creates a view of the the original data, so it's constant time. It'll also reflect changes to the original array and keep the whole original array in memory:

>>> a
array([1, 2, 3, 4, 5])
>>> b = a[::2]         # O(1), constant time
>>> b[:] = 0           # modifying the view changes original array
>>> a                  # original array is modified
array([0, 2, 0, 4, 0])

so if either of the above things are a problem, you can make a copy explicitly:

>>> a
array([1, 2, 3, 4, 5])
>>> b = a[::2].copy()  # explicit copy, O(n)
>>> b[:] = 0           # modifying the copy
>>> a                  # original is intact
array([1, 2, 3, 4, 5])

This isn't constant time, but the result isn't tied to the original array. The copy also contiguous in memory, which can make some operations on it faster.

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])

You can use slicing against the columns

>>> a
array([[1, 2, 3, 4, 5],
       [1, 2, 3, 4, 5],
       [1, 2, 3, 4, 5]])

>>> a[:, ::2]
array([[1, 3, 5],
       [1, 3, 5],
       [1, 3, 5]])

As mentioned by @tobias_k if you want to make an actual copy of this sliced array, you can use numpy.copy to make sure modifications don't affect the original array

>>> np.copy(a[:, ::2])
array([[1, 3, 5],
       [1, 3, 5],
       [1, 3, 5]])