scipy.ndimage.distance_transform_edt#

scipy.ndimage.distance_transform_edt(input, sampling=None, return_distances=True, return_indices=False, distances=None, indices=None)[source]#

Exact Euclidean distance transform.

In addition to the distance transform, the feature transform can be calculated. In this case the index of the closest background element to each foreground element is returned in a separate array.

Parameters

inputarray_like: Input data to transform. Can be any type but will be converted into binary: 1 wherever input equates to True, 0 elsewhere.
samplingfloat, or sequence of float, optional: Spacing of elements along each dimension. If a sequence, must be of length equal to the input rank; if a single number, this is used for all axes. If not specified, a grid spacing of unity is implied.
return_distancesbool, optional: Whether to calculate the distance transform. Default is True.
return_indicesbool, optional: Whether to calculate the feature transform. Default is False.
distancesfloat64 ndarray, optional: An output array to store the calculated distance transform, instead of returning it. return_distances must be True. It must be the same shape as input.
indicesint32 ndarray, optional: An output array to store the calculated feature transform, instead of returning it. return_indicies must be True. Its shape must be (input.ndim,) + input.shape.

Returns

distancesfloat64 ndarray, optional: The calculated distance transform. Returned only when return_distances is True and distances is not supplied. It will have the same shape as the input array.
indicesint32 ndarray, optional: The calculated feature transform. It has an input-shaped array for each dimension of the input. See example below. Returned only when return_indices is True and indices is not supplied.

Notes

The Euclidean distance transform gives values of the Euclidean distance:

              n
y_i = sqrt(sum (x[i]-b[i])**2)
              i

where b[i] is the background point (value 0) with the smallest Euclidean distance to input points x[i], and n is the number of dimensions.

Examples

>>> from scipy import ndimage
>>> a = np.array(([0,1,1,1,1],
...               [0,0,1,1,1],
...               [0,1,1,1,1],
...               [0,1,1,1,0],
...               [0,1,1,0,0]))
>>> ndimage.distance_transform_edt(a)
array([[ 0.    ,  1.    ,  1.4142,  2.2361,  3.    ],
       [ 0.    ,  0.    ,  1.    ,  2.    ,  2.    ],
       [ 0.    ,  1.    ,  1.4142,  1.4142,  1.    ],
       [ 0.    ,  1.    ,  1.4142,  1.    ,  0.    ],
       [ 0.    ,  1.    ,  1.    ,  0.    ,  0.    ]])

With a sampling of 2 units along x, 1 along y:

>>> ndimage.distance_transform_edt(a, sampling=[2,1])
array([[ 0.    ,  1.    ,  2.    ,  2.8284,  3.6056],
       [ 0.    ,  0.    ,  1.    ,  2.    ,  3.    ],
       [ 0.    ,  1.    ,  2.    ,  2.2361,  2.    ],
       [ 0.    ,  1.    ,  2.    ,  1.    ,  0.    ],
       [ 0.    ,  1.    ,  1.    ,  0.    ,  0.    ]])

Asking for indices as well:

>>> edt, inds = ndimage.distance_transform_edt(a, return_indices=True)
>>> inds
array([[[0, 0, 1, 1, 3],
        [1, 1, 1, 1, 3],
        [2, 2, 1, 3, 3],
        [3, 3, 4, 4, 3],
        [4, 4, 4, 4, 4]],
       [[0, 0, 1, 1, 4],
        [0, 1, 1, 1, 4],
        [0, 0, 1, 4, 4],
        [0, 0, 3, 3, 4],
        [0, 0, 3, 3, 4]]])

With arrays provided for inplace outputs:

>>> indices = np.zeros(((np.ndim(a),) + a.shape), dtype=np.int32)
>>> ndimage.distance_transform_edt(a, return_indices=True, indices=indices)
array([[ 0.    ,  1.    ,  1.4142,  2.2361,  3.    ],
       [ 0.    ,  0.    ,  1.    ,  2.    ,  2.    ],
       [ 0.    ,  1.    ,  1.4142,  1.4142,  1.    ],
       [ 0.    ,  1.    ,  1.4142,  1.    ,  0.    ],
       [ 0.    ,  1.    ,  1.    ,  0.    ,  0.    ]])
>>> indices
array([[[0, 0, 1, 1, 3],
        [1, 1, 1, 1, 3],
        [2, 2, 1, 3, 3],
        [3, 3, 4, 4, 3],
        [4, 4, 4, 4, 4]],
       [[0, 0, 1, 1, 4],
        [0, 1, 1, 1, 4],
        [0, 0, 1, 4, 4],
        [0, 0, 3, 3, 4],
        [0, 0, 3, 3, 4]]])