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pythonnumpyscipyinterpolation
John Titor
John Titor

Reputation: 572

Python: Fast discrete interpolation

Given a 4D array that represents a discrete coordinate transformation function such that

arr[x_in, y_in, z_in] = [x_out, y_out, z_out]

I would like to interpolate arr to a grid with more elements (assuming that the samples in arr were initially drawn from a regularly spaced grid of the higher-element cube). I have tried the RegularGridInterpolator from scipy, however this has been rather slow:

import numpy as np
from scipy.interpolate import RegularGridInterpolator
from time import time

target_size   = 32
reduced_size  = 5

small_shape = (reduced_size,reduced_size,reduced_size,3)
cube_small  = np.random.randint(target_size, size=small_shape, dtype=np.uint8)

igrid = 3*[np.linspace(0, target_size-1, reduced_size)]
large_shape = (target_size, target_size, target_size,3)
cube_large  = np.empty(large_shape)

t0 = time()
interpol = RegularGridInterpolator(igrid, cube_small)
for x in np.arange(target_size):
    for y in np.arange(target_size):
        for z in np.arange(target_size):
            cube_large[x,y,z] = interpol([x,y,z])
print(time()-t0)

Are there any algorithms that come to mind that would be better suited for the task? Maybe there is something that could exploit the fact that in this case, I am only interested in the discrete integer values at each point.

Upvotes: 5

Views: 598

Answers (2)

John Titor
John Titor

Reputation: 572

Indeed passing the grid as a whole, as suggested by Crivella seems to be much more effective. I am adding a numpy only version which seems to be about 2 times faster than itertools:

target_size = 256

... # Same as above

def do_new():
    t0 = time()
    grid = np.meshgrid(*3*[np.arange(target_size)], indexing='ij')
    grid = grid[::-1]
    grid = np.stack(grid).T
    ret  = interpol(grid)
    print(f"{time()-t0}")
    return ret

c1 = improved_method()
c2 = do_new()
print((c1==c2).all())

Output:

IMPROVED METHOD: 13.865211009979248
6.268443584442139
True

Upvotes: 0

Crivella
Crivella

Reputation: 1007

I can't really say for the generation of the grid as I am not totally sure about what you are trying to do. But with respect to just improving the efficiency, using a triple for loop instead of making use of broadcasting is massively slowing down your code

import itertools
import numpy as np
from scipy.interpolate import RegularGridInterpolator
from time import time

target_size   = 32
reduced_size  = 5

small_shape = (reduced_size,reduced_size,reduced_size,3)
cube_small  = np.random.randint(target_size, size=small_shape, dtype=np.uint8)


igrid = 3*[np.linspace(0, target_size-1, reduced_size)]


large_shape = (target_size, target_size, target_size,3)
cube_large  = np.empty(large_shape)

def original_method():
    t0 = time()

    interpol = RegularGridInterpolator(igrid, cube_small)
    for x in np.arange(target_size):
        for y in np.arange(target_size):
            for z in np.arange(target_size):
                cube_large[x,y,z] = interpol([x,y,z])

    print('ORIGINAL METHOD: ', time()-t0)
    return cube_large

def improved_method():
    t1 = time()
    interpol = RegularGridInterpolator(igrid, cube_small)

    arr = np.arange(target_size)
    grid = np.array(list(itertools.product(arr, repeat=3)))
    cube_large = interpol(grid).reshape(target_size, target_size, target_size, 3)

    print('IMPROVED METHOD:', time() - t1)

    return cube_large


c1 = original_method()
c2 = improved_method()

print('Is the result the same?  ', np.all(c1 == c2))

OUTPUT

ORIGINAL METHOD:  6.9040000438690186
IMPROVED METHOD: 0.026999950408935547
Is the result the same?   True

Upvotes: 2

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