scipy.ndimage.

binary_fill_holes#

scipy.ndimage.binary_fill_holes(input, structure=None, output=None, origin=0, *, axes=None)[source]#

Fill the holes in binary objects.

Parameters:
inputarray_like

N-D binary array with holes to be filled

structurearray_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.

outputndarray, optional

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

originint, tuple of ints, optional

Position of the structuring element.

axestuple of int or None

The axes over which to apply the filter. If None, input is filtered along all axes. If an origin tuple is provided, its length must match the number of axes.

Returns:
outndarray

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

Examples

>>> from scipy import ndimage
>>> import numpy as np
>>> 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]])
On this page