Python 3D interpolation speedup

问题

I have following code used to interpolate 3D volume data.

Y, X, Z = np.shape(volume)
xs = np.arange(0, X)
ys = np.arange(0, Y)
zs = np.arange(0, Z)

points = list(zip(np.ravel(result[:, :, :, 1]), np.ravel(result[:, :, :, 0]), np.ravel(result[:, :, :, 2])))
interp = interpolate.RegularGridInterpolator((ys, xs, zs), volume,
                                             bounds_error=False, fill_value=0, method='linear')
new_volume = interp(points)
new_volume = np.reshape(new_volume, (Y, X, Z))

This code takes about 37 seconds to execute on 512x512x110 volume (about 29 millions of points), which results in more than one microsecond per voxel (which is unacceptable amount of time for me - what is more it uses 4 cores). Call new_volume=interp(points) takes about 80% of the prodecure time and the list creation almost whole remaining time.

Is there any simple (or even more complex) way to make this computation faster? Or is there any good Python library, which provides faster interpolation? My volume and points change in every call to this prodecure.

回答1:

Here is slightly modified version of your cython solution:

import numpy as np
cimport numpy as np
from libc.math cimport floor
from cython cimport boundscheck, wraparound, nonecheck, cdivision

DTYPE = np.float
ctypedef np.float_t DTYPE_t

@boundscheck(False)
@wraparound(False)
@nonecheck(False)
def interp3D(DTYPE_t[:,:,::1] v, DTYPE_t[:,:,::1] xs, DTYPE_t[:,:,::1] ys, DTYPE_t[:,:,::1] zs):

    cdef int X, Y, Z
    X,Y,Z = v.shape[0], v.shape[1], v.shape[2]
    cdef np.ndarray[DTYPE_t, ndim=3] interpolated = np.zeros((X, Y, Z), dtype=DTYPE)

    _interp3D(&v[0,0,0], &xs[0,0,0], &ys[0,0,0], &zs[0,0,0], &interpolated[0,0,0], X, Y, Z)
    return interpolated


@cdivision(True)
cdef inline void _interp3D(DTYPE_t *v, DTYPE_t *x_points, DTYPE_t *y_points, DTYPE_t *z_points, 
               DTYPE_t *result, int X, int Y, int Z):

    cdef:
        int i, x0, x1, y0, y1, z0, z1, dim
        DTYPE_t x, y, z, xd, yd, zd, c00, c01, c10, c11, c0, c1, c

    dim = X*Y*Z

    for i in range(dim):
        x = x_points[i]
        y = y_points[i]
        z = z_points[i]

        x0 = <int>floor(x)
        x1 = x0 + 1
        y0 = <int>floor(y)
        y1 = y0 + 1
        z0 = <int>floor(z)
        z1 = z0 + 1

        xd = (x-x0)/(x1-x0)
        yd = (y-y0)/(y1-y0)
        zd = (z-z0)/(z1-z0)

        if x0 >= 0 and y0 >= 0 and z0 >= 0:
            c00 = v[Y*Z*x0+Z*y0+z0]*(1-xd) + v[Y*Z*x1+Z*y0+z0]*xd
            c01 = v[Y*Z*x0+Z*y0+z1]*(1-xd) + v[Y*Z*x1+Z*y0+z1]*xd
            c10 = v[Y*Z*x0+Z*y1+z0]*(1-xd) + v[Y*Z*x1+Z*y1+z0]*xd
            c11 = v[Y*Z*x0+Z*y1+z1]*(1-xd) + v[Y*Z*x1+Z*y1+z1]*xd

            c0 = c00*(1-yd) + c10*yd
            c1 = c01*(1-yd) + c11*yd

            c = c0*(1-zd) + c1*zd

        else:
            c = 0

        result[i] = c

The results are still identical to yours. With a random grid data of 60x60x60 I obtain the following timings:

SciPy's solution:                982ms
Your cython solution:            24.7ms
Above modified cython solution:  8.17ms

So its nearly 4 times faster than your cython solution. Note that

Cython performs bounds checking by default on arrays and if you want that feature then turn on boundschecking remove @boundscheck(False).
In the above solution the arrays are required to be C-contiguous
If you want a parallel variant of the above code, replace range instead of prange in your for loop.

Hope this helps.

回答2:

I used Cython to accelerate this and implemented following code:

import numpy as np
cimport numpy as np
np.import_array()
from libc.math cimport ceil, floor

DTYPE = np.float
ctypedef np.float_t DTYPE_t

def interp3(np.ndarray[DTYPE_t, ndim=3] x_grid, np.ndarray[DTYPE_t, ndim=3] y_grid,
    np.ndarray[DTYPE_t, ndim=3] z_grid, np.ndarray[DTYPE_t, ndim=3] v,
    np.ndarray[DTYPE_t, ndim=3] xs, np.ndarray[DTYPE_t, ndim=3] ys, 
    np.ndarray[DTYPE_t, ndim=3] zs):

    cdef int i
    cdef float x
    cdef float y
    cdef float z
    cdef int x0
    cdef int x1
    cdef int y0
    cdef int y1
    cdef int z0
    cdef int z1
    cdef float xd
    cdef float yd
    cdef float zd
    cdef float c00
    cdef float c01
    cdef float c10
    cdef float c11
    cdef float c0
    cdef float c1
    cdef float c
    cdef int X
    cdef int Y
    cdef int Z

    X, Y, Z = np.shape(x_grid)

    cdef np.ndarray[DTYPE_t, ndim=1] x_points = np.ravel(xs)
    cdef np.ndarray[DTYPE_t, ndim=1] y_points = np.ravel(ys)
    cdef np.ndarray[DTYPE_t, ndim=1] z_points = np.ravel(zs)
    cdef np.ndarray[DTYPE_t, ndim=1] result = np.empty((len(x_points)), dtype=DTYPE)

    for i in range(len(x_points)):
        x = x_points[i]
        y = y_points[i]
        z = z_points[i]

        x0 = int(floor(x))
        x1 = x0 + 1
        y0 = int(floor(y))
        y1 = y0 + 1
        z0 = int(floor(z))
        z1 = z0 + 1

        xd = (x-x0)/(x1-x0)
        yd = (y-y0)/(y1-y0)
        zd = (z-z0)/(z1-z0)

        try:
            assert x0 >= 0 and y0 >= 0 and z0 >= 0
            c00 = v[x0, y0, z0]*(1-xd) + v[x1, y0, z0]*xd
            c01 = v[x0, y0, z1]*(1-xd) + v[x1, y0, z1]*xd
            c10 = v[x0, y1, z0]*(1-xd) + v[x1, y1, z0]*xd
            c11 = v[x0, y1, z1]*(1-xd) + v[x1, y1, z1]*xd

            c0 = c00*(1-yd) + c10*yd
            c1 = c01*(1-yd) + c11*yd

            c = c0*(1-zd) + c1*zd
        except:
            c = 0

        result[i] = c

    cdef np.ndarray[DTYPE_t, ndim=3] interpolated = np.zeros((X, Y, Z), dtype=DTYPE)
    interpolated = np.reshape(result, (X, Y, Z))
    return interpolated

It is my first time with Cython, so have following questions:

How can I optimize this further?
Is there any easy way to avoid try and assert statements to check array bounds? Trying to ensure bounds with min/max combinations is slower than this try/assert approach

Currently, it is around 8x faster than original code posted above.

来源：https://stackoverflow.com/questions/41220617/python-3d-interpolation-speedup

标签

python

performance

numpy

interpolation

cython