Speedup scipy griddata for multiple interpolations between two irregular grids

Question

I have several values that are defined on the same irregular grid (x, y, z) that I want to interpolate onto a new grid (x1, y1, z1). i.e., I have <

Answer 1

Great thanks to Jaime for his solution (even if I don't really understand how the barycentric computation is done ...)

Here you will find an example adapted from his case in 2D :

import scipy.interpolate as spint
import scipy.spatial.qhull as qhull
import numpy as np

def interp_weights(xy, uv,d=2):
    tri = qhull.Delaunay(xy)
    simplex = tri.find_simplex(uv)
    vertices = np.take(tri.simplices, simplex, axis=0)
    temp = np.take(tri.transform, simplex, axis=0)
    delta = uv - temp[:, d]
    bary = np.einsum('njk,nk->nj', temp[:, :d, :], delta)
    return vertices, np.hstack((bary, 1 - bary.sum(axis=1, keepdims=True)))

def interpolate(values, vtx, wts):
    return np.einsum('nj,nj->n', np.take(values, vtx), wts)

m, n = 101,201
mi, ni = 1001,2001

[Y,X]=np.meshgrid(np.linspace(0,1,n),np.linspace(0,2,m))
[Yi,Xi]=np.meshgrid(np.linspace(0,1,ni),np.linspace(0,2,mi))

xy=np.zeros([X.shape[0]*X.shape[1],2])
xy[:,0]=Y.flatten()
xy[:,1]=X.flatten()
uv=np.zeros([Xi.shape[0]*Xi.shape[1],2])
uv[:,0]=Yi.flatten()
uv[:,1]=Xi.flatten()

values=np.cos(2*X)*np.cos(2*Y)

#Computed once and for all !
vtx, wts = interp_weights(xy, uv)
valuesi=interpolate(values.flatten(), vtx, wts)
valuesi=valuesi.reshape(Xi.shape[0],Xi.shape[1])
print "interpolation error: ",np.mean(valuesi-np.cos(2*Xi)*np.cos(2*Yi))  
print "interpolation uncertainty: ",np.std(valuesi-np.cos(2*Xi)*np.cos(2*Yi))  

It is possible to applied image transformation such as image mapping with a udge speed-up

You can't use the same function definition as the new coordinates will change at every iteration but you can compute triangulation Once for all.

import scipy.interpolate as spint
import scipy.spatial.qhull as qhull
import numpy as np
import time

# Definition of the fast  interpolation process. May be the Tirangulation process can be removed !!
def interp_tri(xy):
    tri = qhull.Delaunay(xy)
    return tri


def interpolate(values, tri,uv,d=2):
    simplex = tri.find_simplex(uv)
    vertices = np.take(tri.simplices, simplex, axis=0)
    temp = np.take(tri.transform, simplex, axis=0)
    delta = uv- temp[:, d]
    bary = np.einsum('njk,nk->nj', temp[:, :d, :], delta)  
    return np.einsum('nj,nj->n', np.take(values, vertices),  np.hstack((bary, 1.0 - bary.sum(axis=1, keepdims=True))))

m, n = 101,201
mi, ni = 101,201

[Y,X]=np.meshgrid(np.linspace(0,1,n),np.linspace(0,2,m))
[Yi,Xi]=np.meshgrid(np.linspace(0,1,ni),np.linspace(0,2,mi))

xy=np.zeros([X.shape[0]*X.shape[1],2])
xy[:,1]=Y.flatten()
xy[:,0]=X.flatten()
uv=np.zeros([Xi.shape[0]*Xi.shape[1],2])
# creation of a displacement field
uv[:,1]=0.5*Yi.flatten()+0.4
uv[:,0]=1.5*Xi.flatten()-0.7
values=np.zeros_like(X)
values[50:70,90:150]=100.

#Computed once and for all !
tri = interp_tri(xy)
t0=time.time()
for i in range(0,100):
  values_interp_Qhull=interpolate(values.flatten(),tri,uv,2).reshape(Xi.shape[0],Xi.shape[1])
t_q=(time.time()-t0)/100

t0=time.time()
values_interp_griddata=spint.griddata(xy,values.flatten(),uv,fill_value=0).reshape(values.shape[0],values.shape[1])
t_g=time.time()-t0

print "Speed-up:", t_g/t_q
print "Mean error: ",(values_interp_Qhull-values_interp_griddata).mean()
print "Standard deviation: ",(values_interp_Qhull-values_interp_griddata).std()

On my laptop the speed-up is between 20 and 40x !

Hope that can help someone