interpolate 3D volume with numpy and or scipy

Question

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

Answer 1

The exact equivalent to MATLAB's interp3 would be using scipy's interpn for one-off interpolation:

import numpy as np
from scipy.interpolate import interpn

Vi = interpn((x,y,z), V, np.array([xi,yi,zi]).T)

The default method for both MATLAB and scipy is linear interpolation, and this can be changed with the method argument. Note that only linear and nearest-neighbor interpolation is supported by interpn for 3 dimensions and above, unlike MATLAB which supports cubic and spline interpolation as well.

When making multiple interpolation calls on the same grid it is preferable to use the interpolation object RegularGridInterpolator, as in the accepted answer above. interpn uses RegularGridInterpolator internally.

Answer 2

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.

Answer 3

Basically, ndimage.map_coordinates works in "index" coordinates (a.k.a. "voxel" or "pixel" coordinates). The interface to it seems a bit clunky at first, but it does give you a lot of flexibility.

If you want to specify the interpolated coordinates similar to matlab's interp3, then you'll need to convert your intput coordinates into "index" coordinates.

There's also the additional wrinkle that map_coordinates always preserves the dtype of the input array in the output. If you interpolate an integer array, you'll get integer output, which may or may not be what you want. For the code snippet below, I'll assume that you always want floating point output. (If you don't, it's actually simpler.)

I'll try to add more explanation later tonight (this is rather dense code).

All in all, the interp3 function I have is more complex than it may need to be for your exact purposes. Howver, it should more or less replicate the behavior of interp3 as I remember it (ignoring the "zooming" functionality of interp3(data, zoom_factor), which scipy.ndimage.zoom handles.)

import numpy as np
from scipy.ndimage import map_coordinates

def main():
    data = np.arange(5*4*3).reshape(5,4,3)

    x = np.linspace(5, 10, data.shape[0])
    y = np.linspace(10, 20, data.shape[1])
    z = np.linspace(-100, 0, data.shape[2])

    # Interpolate at a single point
    print interp3(x, y, z, data, 7.5, 13.2, -27)

    # Interpolate a region of the x-y plane at z=-25
    xi, yi = np.mgrid[6:8:10j, 13:18:10j]
    print interp3(x, y, z, data, xi, yi, -25 * np.ones_like(xi))

def interp3(x, y, z, v, xi, yi, zi, **kwargs):
    """Sample a 3D array "v" with pixel corner locations at "x","y","z" at the
    points in "xi", "yi", "zi" using linear interpolation. Additional kwargs
    are passed on to ``scipy.ndimage.map_coordinates``."""
    def index_coords(corner_locs, interp_locs):
        index = np.arange(len(corner_locs))
        if np.all(np.diff(corner_locs) < 0):
            corner_locs, index = corner_locs[::-1], index[::-1]
        return np.interp(interp_locs, corner_locs, index)

    orig_shape = np.asarray(xi).shape
    xi, yi, zi = np.atleast_1d(xi, yi, zi)
    for arr in [xi, yi, zi]:
        arr.shape = -1

    output = np.empty(xi.shape, dtype=float)
    coords = [index_coords(*item) for item in zip([x, y, z], [xi, yi, zi])]

    map_coordinates(v, coords, order=1, output=output, **kwargs)

    return output.reshape(orig_shape)

main()