scipy.ndimage.binary_fill_holes¶

scipy.ndimage.
binary_fill_holes
(input, structure=None, output=None, origin=0)[source]¶ Fill the holes in binary objects.
Parameters:  input : array_like
ndimensional binary array with holes to be filled
 structure : array_like, optional
Structuring element used in the computation; largesize 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
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
[1] 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]])