You are looking for scipy.ndimage.morphology.distance_transform_edt
. It operates on a binary array and computes euclidean distances on each TRUE
position to the nearest background FALSE
position. In our case, since we want to find out distances from nearest 0s, so the background is 0
. Now, under the hoods, it converts the input to a binary array assuming 0 as the background, so we can just use it with the default parameters. Hence, it would be as simple as -
In [179]: a
Out[179]:
array([[0, 0, 1, 2, 5, 2, 1],
[0, 0, 2, 3, 5, 2, 1],
[0, 1, 1, 2, 5, 4, 1],
[1, 1, 1, 2, 5, 4, 0]])
In [180]: from scipy import ndimage
In [181]: ndimage.distance_transform_edt(a)
Out[181]:
array([[0. , 0. , 1. , 2. , 3. , 3.16, 3. ],
[0. , 0. , 1. , 2. , 2.83, 2.24, 2. ],
[0. , 1. , 1.41, 2.24, 2.24, 1.41, 1. ],
[1. , 1.41, 2.24, 2.83, 2. , 1. , 0. ]])
Solving for generic case
Now, let's say we want to find out distances from nearest 1s
, then it would be -
In [183]: background = 1 # element from which distances are to be computed
# compare this with original array, a to verify
In [184]: ndimage.distance_transform_edt(a!=background)
Out[184]:
array([[2. , 1. , 0. , 1. , 2. , 1. , 0. ],
[1.41, 1. , 1. , 1.41, 2. , 1. , 0. ],
[1. , 0. , 0. , 1. , 2. , 1. , 0. ],
[0. , 0. , 0. , 1. , 2. , 1.41, 1. ]])