[英]Python 3D interpolation speedup
I have following code used to interpolate 3D volume data. 我有以下用于插入3D体积数据的代码。
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). 此代码在512x512x110卷(大约2900万个点)上执行大约需要37秒,这导致每个体素超过1微秒(这对我来说是不可接受的时间 - 更多的是它使用4个核心)。 Call
new_volume=interp(points)
takes about 80% of the prodecure time and the list creation almost whole remaining time. 调用
new_volume=interp(points)
需要大约80%的prodecure时间和列表创建几乎整个剩余时间。
Is there any simple (or even more complex) way to make this computation faster? 是否有任何简单(甚至更复杂)的方法可以使这种计算更快? Or is there any good Python library, which provides faster interpolation?
或者是否有任何优秀的Python库,它提供更快的插值? My volume and points change in every call to this prodecure.
每次打电话给我这个职业,我的数量和积分都会发生变化。
Here is slightly modified version of your cython
solution: 这是你的
cython
解决方案的略微修改版本:
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. 结果仍然与您的相同。 With a random grid data of
60x60x60
I obtain the following timings: 使用
60x60x60
的随机网格数据,我获得以下时间:
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. 所以它比你的
cython
解决方案快近4倍。 Note that 注意
@boundscheck(False)
. @boundscheck(False)
。 range
instead of prange
in your for loop
. prange
在for loop
替换range
而不是prange
。 Hope this helps. 希望这可以帮助。
I used Cython to accelerate this and implemented following code: 我使用Cython来加速这个并实现以下代码:
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: 这是我第一次使用Cython,所以有以下问题:
How can I optimize this further? 我该如何进一步优化?
Is there any easy way to avoid try and assert statements to check array bounds? 有没有简单的方法来避免try和assert语句来检查数组边界? Trying to ensure bounds with min/max combinations is slower than this try/assert approach
尝试确保与最小/最大组合的边界比这个try / assert方法慢
Currently, it is around 8x faster than original code posted above. 目前,它比上面发布的原始代码快约8倍。
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.