4

Another, similar post called Flood Fill in Python is a very general question on flood fill and the answer only contains a broad pseudo code example. I'm look for an explicit solution with numpy or scipy.

Let's take this array for example:

a = np.array([
    [0, 1, 1, 1, 1, 0],
    [0, 0, 1, 2, 1, 1],
    [0, 1, 1, 1, 1, 0]
])

For selecting element 0, 0 and flood fill with value 3, I'd expect:

[
    [3, 1, 1, 1, 1, 0],
    [3, 3, 1, 2, 1, 1],
    [3, 1, 1, 1, 1, 0]
]

For selecting element 0, 1 and flood fill with value 3, I'd expect:

[
    [0, 3, 3, 3, 3, 0],
    [0, 0, 3, 2, 3, 3],
    [0, 3, 3, 3, 3, 0]
]

For selecting element 0, 5 and flood fill with value 3, I'd expect:

[
    [0, 1, 1, 1, 1, 3],
    [0, 0, 1, 2, 1, 1],
    [0, 1, 1, 1, 1, 0]
]

This should be a fairly basic operation, no? Which numpy or scipy method am I overlooking?

finefoot
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  • Possible duplicate of [Flood Fill in Python](https://stackoverflow.com/questions/19839947/flood-fill-in-python) – Stop harming Monica May 26 '19 at 16:00
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    @Goyo That's a pseudo-code and don't see that working with array without wholesome changes to the accepted solution. How do you propose to use the accepted solution there to solve it here? – Divakar May 26 '19 at 17:41

1 Answers1

10

Approach #1

Module scikit-image offers the built-in to do the same with skimage.segmentation.flood_fill -

from skimage.morphology import flood_fill

flood_fill(image, (x, y), newval)

Sample runs -

In [17]: a
Out[17]: 
array([[0, 1, 1, 1, 1, 0],
       [0, 0, 1, 2, 1, 1],
       [0, 1, 1, 1, 1, 0]])

In [18]: flood_fill(a, (0, 0), 3)
Out[18]: 
array([[3, 1, 1, 1, 1, 0],
       [3, 3, 1, 2, 1, 1],
       [3, 1, 1, 1, 1, 0]])

In [19]: flood_fill(a, (0, 1), 3)
Out[19]: 
array([[0, 3, 3, 3, 3, 0],
       [0, 0, 3, 2, 3, 3],
       [0, 3, 3, 3, 3, 0]])

In [20]: flood_fill(a, (0, 5), 3)
Out[20]: 
array([[0, 1, 1, 1, 1, 3],
       [0, 0, 1, 2, 1, 1],
       [0, 1, 1, 1, 1, 0]])

Approach #2

We can use skimage.measure.label with some array-masking -

from skimage.measure import label

def floodfill_by_xy(a,xy,newval):
    x,y = xy
    l = label(a==a[x,y])
    a[l==l[x,y]] = newval
    return a

To make use of SciPy based label function - scipy.ndimage.measurements.label, it would mostly be the same -

from scipy.ndimage.measurements import label

def floodfill_by_xy_scipy(a,xy,newval):
    x,y = xy
    l = label(a==a[x,y])[0]
    a[l==l[x,y]] = newval
    return a

Note : These would work as in-situ edits.

Divakar
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