scipy.ndimage.binary_fill_holes

scipy.ndimage.binary_fill_holes(input, structure=None, output=None, origin=0)

Fill the holes in binary objects.

Parameters:

input : array_like

n-dimensional binary array with holes to be filled

structure : array_like, optional

Structuring element used in the computation; large-size elements make computations faster but may miss holes separated from the background by thin regions. The default element (with a square connectivity equal to one) yields the intuitive result where all holes in the input have been filled.

output : ndarray, optional

Array of the same shape as input, into which the output is placed. By default, a new array is created.

origin : int, tuple of ints, optional

Position of the structuring element.

Returns:

out : ndarray

Transformation of the initial image input where holes have been filled.

See also

binary_dilation, binary_propagation, label

Notes

The algorithm used in this function consists in invading the complementary of the shapes in input from the outer boundary of the image, using binary dilations. Holes are not connected to the boundary and are therefore not invaded. The result is the complementary subset of the invaded region.

References

[R139]

http://en.wikipedia.org/wiki/Mathematical_morphology

Examples

>>> from scipy import ndimage
>>> a = np.zeros((5, 5), dtype=int)
>>> a[1:4, 1:4] = 1
>>> a[2,2] = 0
>>> a
array([[0, 0, 0, 0, 0],
       [0, 1, 1, 1, 0],
       [0, 1, 0, 1, 0],
       [0, 1, 1, 1, 0],
       [0, 0, 0, 0, 0]])
>>> ndimage.binary_fill_holes(a).astype(int)
array([[0, 0, 0, 0, 0],
       [0, 1, 1, 1, 0],
       [0, 1, 1, 1, 0],
       [0, 1, 1, 1, 0],
       [0, 0, 0, 0, 0]])
>>> # Too big structuring element
>>> ndimage.binary_fill_holes(a, structure=np.ones((5,5))).astype(int)
array([[0, 0, 0, 0, 0],
       [0, 1, 1, 1, 0],
       [0, 1, 0, 1, 0],
       [0, 1, 1, 1, 0],
       [0, 0, 0, 0, 0]])