interpolate 3D volume with numpy and or scipy
3d, interpolation, numpy, python, scipy
Solution
In scipy 0.14 or later, there is a new function `scipy.interpolate.RegularGridInterpolator` which closely resembles `interp3`.
The MATLAB command `Vi = interp3(x,y,z,V,xi,yi,zi)` would translate to something like:
from numpy import array
from scipy.interpolate import RegularGridInterpolator as rgi
my_interpolating_function = rgi((x,y,z), V)
Vi = my_interpolating_function(array([xi,yi,zi]).T)
Here is a full example demonstrating both; it will help you understand the exact differences...
MATLAB CODE:
x = linspace(1,4,11);
y = linspace(4,7,22);
z = linspace(7,9,33);
V = zeros(22,11,33);
for i=1:11
for j=1:22
for k=1:33
V(j,i,k) = 100*x(i) + 10*y(j) + z(k);
end
end
end
xq = [2,3];
yq = [6,5];
zq = [8,7];
Vi = interp3(x,y,z,V,xq,yq,zq);
The result is `Vi=[268 357]` which is indeed the value at those two points `(2,6,8)` and `(3,5,7)`.
SCIPY CODE:
from scipy.interpolate import RegularGridInterpolator
from numpy import linspace, zeros, array
x = linspace(1,4,11)
y = linspace(4,7,22)
z = linspace(7,9,33)
V = zeros((11,22,33))
for i in range(11):
for j in range(22):
for k in range(33):
V[i,j,k] = 100*x[i] + 10*y[j] + z[k]
fn = RegularGridInterpolator((x,y,z), V)
pts = array([[2,6,8],[3,5,7]])
print(fn(pts))
Again it's `[268,357]`. So you see some slight differences: Scipy uses x,y,z index order while MATLAB uses y,x,z (strangely); In Scipy you define a function in a separate step and when you call it, the coordinates are grouped like (x1,y1,z1),(x2,y2,z2),... while matlab uses (x1,x2,...),(y1,y2,...),(z1,z2,...).
Other than that, the two are similar and equally easy to use.
Problem
I am extremely frustrated because after several hours I can't seem to be able to do a seemingly easy 3D interpolation in python. In Matlab all I had to do was ``` Vi = interp3(x,y,z,V,xi,yi,zi) ``` What is the exact equivalent of this using scipy's ndimage.map_coordinate or other numpy methods? Thanks