https://docs.scipy.org/doc/scipy-0.14.0/reference/generated/scipy.ndimage.morphology.distance_transform_edt.html
I\'m having trouble understanding how the Euclidean dist
It is not clear from the docstring, but distance_transform_edt computes the distance from non-zero (i.e. non-background) points to the nearest zero (i.e. background) point.
For example:
In [42]: x
Out[42]:
array([[0, 0, 0, 0, 0, 1, 1, 1],
[0, 1, 1, 1, 0, 1, 1, 1],
[0, 1, 1, 1, 0, 1, 1, 1],
[0, 0, 1, 1, 0, 0, 0, 1]])
In [43]: np.set_printoptions(precision=3) # Easier to read the result with fewer digits.
In [44]: distance_transform_edt(x)
Out[44]:
array([[ 0. , 0. , 0. , 0. , 0. , 1. , 2. , 3. ],
[ 0. , 1. , 1. , 1. , 0. , 1. , 2. , 2.236],
[ 0. , 1. , 1.414, 1. , 0. , 1. , 1. , 1.414],
[ 0. , 0. , 1. , 1. , 0. , 0. , 0. , 1. ]])
You can get the equivalent of Matlab's bwdist(a)
by applying distance_transform_edt()
to np.logical_not(a)
(i.e. invert the foreground and background):
In [71]: a
Out[71]:
array([[ 0., 0., 0., 0., 0.],
[ 0., 1., 0., 0., 0.],
[ 0., 0., 0., 0., 0.],
[ 0., 0., 0., 1., 0.],
[ 0., 0., 0., 0., 0.]])
In [72]: distance_transform_edt(np.logical_not(a))
Out[72]:
array([[ 1.414, 1. , 1.414, 2.236, 3.162],
[ 1. , 0. , 1. , 2. , 2.236],
[ 1.414, 1. , 1.414, 1. , 1.414],
[ 2.236, 2. , 1. , 0. , 1. ],
[ 3.162, 2.236, 1.414, 1. , 1.414]])
Warren has already explained how distance_transform_edt
works.
In your case,you could change sampling units along x and y
ndimage.distance_transform_edt(a)
array([[ 0., 0., 0., 0., 0.],
[ 0., 1., 0., 0., 0.],
[ 0., 0., 0., 0., 0.],
[ 0., 0., 0., 1., 0.],
[ 0., 0., 0., 0., 0.]])
But
>>> ndimage.distance_transform_edt(a, sampling=[2,2])
array([[ 0., 0., 0., 0., 0.],
[ 0., 2., 0., 0., 0.],
[ 0., 0., 0., 0., 0.],
[ 0., 0., 0., 2., 0.],
[ 0., 0., 0., 0., 0.]])
Or
ndimage.distance_transform_edt(a, sampling=[3,3])
array([[ 0., 0., 0., 0., 0.],
[ 0., 3., 0., 0., 0.],
[ 0., 0., 0., 0., 0.],
[ 0., 0., 0., 3., 0.],
[ 0., 0., 0., 0., 0.]])