Extend numpy mask by n cells to the right for each bad value, efficiently

Question

Let\'s say I have a length 30 array with 4 bad values in it. I want to create a mask for those bad values, but since I will be using rolling window functions, I\'d also like a f

Answer 1

This could also be considered a morphological dilation problem, using here the scipy.ndimage.binary_dilation:

def dilation(a, n):
    m = np.isnan(a)
    s = np.full(n, True, bool)
    return ndimage.binary_dilation(m, structure=s, origin=-(n//2))

Note on origin: this argument ensures the structure (I would call it a kernel) starts off a bit to the left of the input (your mask m). Normally the value at out[i] would be the dilation with the center of structure (which would be structure[n//2]) at in[i], but you want the structure[0] to be at in[i].

You can also do this with a kernel that is padded on the left with Falses, which is what would be required if you used the binary_dilation from scikit-image:

def dilation_skimage(a, n):
    m = np.isnan(a)
    s = np.zeros(2*n - n%2, bool)
    s[-n:] = True
    return skimage.morphology.binary_dilation(m, selem=s)

Timing doesn't seem to change too much between the two:

dilation_scipy
small:    10 loops, best of 3: 47.9 ms per loop
large: 10000 loops, best of 3: 88.9 µs per loop

dilation_skimage
small:    10 loops, best of 3: 47.0 ms per loop
large: 10000 loops, best of 3: 91.1 µs per loop