我有一个3D数组，我需要在一个轴（最后一个维度）上插值。我们说y.shape = (nx, ny, nz)，我想在每个nz的{{1}}中进行插值。但是，我想在每个(nx, ny)中插入不同的值。
这里有一些代码可以举例说明。如果我想插入单个值，比如说[i, j]，我会像这样使用new_z
scipy.interpolate.interp1d但是，对于这个问题，我真正想要的是为每个# y is a 3D ndarray
# x is a 1D ndarray with the abcissa values
# new_z is a number
f = scipy.interpolate.interp1d(x, y, axis=-1, kind='linear')
result = f(new_z)
插入不同的new_z。所以我这样做：
y[i, j]不幸的是，对于多个循环，这变得效率低且速度慢。有没有更好的方法来进行这种插值？线性插值就足够了。一种可能性是在Cython中实现这一点，但我试图避免这种情况，因为我希望能够灵活地更改为三次插值，而不想在Cython中手动完成。
答案 0 :(得分：7)
要加快高阶插值，您只能调用interp1d()一次，然后使用_spline模块中的_bspleval()属性和低级函数_fitpack。这是代码：
from scipy.interpolate import interp1d
import numpy as np
nx, ny, nz = 30, 40, 50
x = np.arange(0, nz, 1.0)
y = np.random.randn(nx, ny, nz)
new_x = np.random.random_integers(1, (nz-1)*10, size=(nx, ny))/10.0
def original_interpolation(x, y, new_x):
result = np.empty(y.shape[:-1])
for i in xrange(nx):
for j in xrange(ny):
f = interp1d(x, y[i, j], axis=-1, kind=3)
result[i, j] = f(new_x[i, j])
return result
def fast_interpolation(x, y, new_x):
from scipy.interpolate._fitpack import _bspleval
f = interp1d(x, y, axis=-1, kind=3)
xj,cvals,k = f._spline
result = np.empty_like(new_x)
for (i, j), value in np.ndenumerate(new_x):
result[i, j] = _bspleval(value, x, cvals[:, i, j], k, 0)
return result
r1 = original_interpolation(x, y, new_x)
r2 = fast_interpolation(x, y, new_x)
>>> np.allclose(r1, r2)
True
%timeit original_interpolation(x, y, new_x)
%timeit fast_interpolation(x, y, new_x)
1 loops, best of 3: 3.78 s per loop
100 loops, best of 3: 15.4 ms per loop
答案 1 :(得分：4)
我认为interp1d有一种快速执行此操作的方法，因此您无法避免此循环。
Cython你可能仍然可以通过使用np.searchsorted编码线性插值来避免，类似这样（未经测试）：
def interp3d(x, y, new_x):
assert x.ndim == 1 and y.ndim == 3 and new_x.ndim == 2
assert y.shape[:2] == new_x.shape and x.shape == y.shape[2:]
nx, ny = y.shape[:2]
new_x = new_x.ravel()
j = np.arange(len(new_x))
k = np.searchsorted(x, new_x).clip(1, len(x) - 1)
y = y.reshape(-1, x.shape[0])
p = (new_x - x[k-1]) / (x[k] - x[k-1])
result = (1 - p) * y[j,k-1] + p * y[j,k]
return result.reshape(nx, ny)
但是对于三次插值没有帮助。
编辑：使其成为一个功能并修复了一个错误。一些时间与Cython（500x500x500网格）：
In [58]: %timeit interp3d(x, y, new_x)
10 loops, best of 3: 82.7 ms per loop
In [59]: %timeit cyfile.interp3d(x, y, new_x)
10 loops, best of 3: 86.3 ms per loop
In [60]: abs(interp3d(x, y, new_x) - cyfile.interp3d(x, y, new_x)).max()
Out[60]: 2.2204460492503131e-16
尽管如此，人们可以争辩说Cython代码更容易阅读。
答案 2 :(得分：3)
由于上面的numpy建议花了太长时间，我可以等一下这里是cython版本供将来参考。从一些松散的基准测试来看，它的速度提高了大约3000倍（授予它，它只是线性插值，并没有像interp1d那么多，但它可以用于此目的。）
import numpy as N
cimport numpy as N
cimport cython
DTYPEf = N.float64
ctypedef N.float64_t DTYPEf_t
@cython.boundscheck(False) # turn of bounds-checking for entire function
@cython.wraparound(False) # turn of bounds-checking for entire function
cpdef interp3d(N.ndarray[DTYPEf_t, ndim=1] x, N.ndarray[DTYPEf_t, ndim=3] y,
N.ndarray[DTYPEf_t, ndim=2] new_x):
"""
interp3d(x, y, new_x)
Performs linear interpolation over the last dimension of a 3D array,
according to new values from a 2D array new_x. Thus, interpolate
y[i, j, :] for new_x[i, j].
Parameters
----------
x : 1-D ndarray (double type)
Array containg the x (abcissa) values. Must be monotonically
increasing.
y : 3-D ndarray (double type)
Array containing the y values to interpolate.
x_new: 2-D ndarray (double type)
Array with new abcissas to interpolate.
Returns
-------
new_y : 3-D ndarray
Interpolated values.
"""
cdef int nx = y.shape[0]
cdef int ny = y.shape[1]
cdef int nz = y.shape[2]
cdef int i, j, k
cdef N.ndarray[DTYPEf_t, ndim=2] new_y = N.zeros((nx, ny), dtype=DTYPEf)
for i in range(nx):
for j in range(ny):
for k in range(1, nz):
if x[k] > new_x[i, j]:
new_y[i, j] = (y[i, j, k] - y[i, j, k - 1]) * \
(new_x[i, j] - x[k-1]) / (x[k] - x[k - 1]) + y[i, j, k - 1]
break
return new_y
答案 3 :(得分：3)
以@pv.'s answer为基础，并对内循环进行矢量化，以下内容提供了显着的加速（编辑：将昂贵的numpy.tile更改为使用numpy.lib.stride_tricks.as_strided）：
import numpy
from scipy import interpolate
nx = 30
ny = 40
nz = 50
y = numpy.random.randn(nx, ny, nz)
x = numpy.float64(numpy.arange(0, nz))
# We select some locations in the range [0.1, nz-0.1]
new_z = numpy.random.random_integers(1, (nz-1)*10, size=(nx, ny))/10.0
# y is a 3D ndarray
# x is a 1D ndarray with the abcissa values
# new_z is a 2D array
def original_interpolation():
result = numpy.empty(y.shape[:-1])
for i in range(nx):
for j in range(ny):
f = interpolate.interp1d(x, y[i, j], axis=-1, kind='linear')
result[i, j] = f(new_z[i, j])
return result
grid_x, grid_y = numpy.mgrid[0:nx, 0:ny]
def faster_interpolation():
flat_new_z = new_z.ravel()
k = numpy.searchsorted(x, flat_new_z)
k = k.reshape(nx, ny)
lower_index = [grid_x, grid_y, k-1]
upper_index = [grid_x, grid_y, k]
tiled_x = numpy.lib.stride_tricks.as_strided(x, shape=(nx, ny, nz),
strides=(0, 0, x.itemsize))
z_upper = tiled_x[upper_index]
z_lower = tiled_x[lower_index]
z_step = z_upper - z_lower
z_delta = new_z - z_lower
y_lower = y[lower_index]
result = y_lower + z_delta * (y[upper_index] - y_lower)/z_step
return result
# both should be the same (giving a small difference)
print numpy.max(
numpy.abs(original_interpolation() - faster_interpolation()))
在我的机器上提供以下时间：
In [8]: timeit foo.original_interpolation()
10 loops, best of 3: 102 ms per loop
In [9]: timeit foo.faster_interpolation()
1000 loops, best of 3: 564 us per loop
转到nx = 300，ny = 300和nz = 500，可获得130倍的加速：
In [2]: timeit original_interpolation()
1 loops, best of 3: 8.27 s per loop
In [3]: timeit faster_interpolation()
10 loops, best of 3: 60.1 ms per loop
你需要编写自己的三次插值算法，但它不应该那么难。
答案 4 :(得分：2)
您可以使用map_coordinates：
from numpy import random, meshgrid, arange
from scipy.ndimage import map_coordinates
(nx, ny, nz) = (4, 5, 6)
# some random array
A = random.rand(nx, ny, nz)
# random floating-point indices in [0, nz-1]
Z = random.rand(nx, ny)*(nz-1)
# regular integer indices of shape (nx,ny)
X, Y = meshgrid(arange(nx), arange(ny), indexing='ij')
coords = (X, Y, Z) # X, Y, and Z are of shape (nx, ny)
print map_coordinates(A, coords, order=1, cval=-999.)
答案 5 :(得分：2)
虽然有几个不错的答案， 他们仍在固定的500长阵列中进行250k插值：
j250k = np.searchsorted( X500, X250k ) # indices in [0, 500)
这可以用LUT，LookUp Table加速，比如5k插槽：
lut = np.interp( np.arange(5000), X500, np.arange(500) ).round().astype(int)
xscale = (X - X.min()) * (5000 - 1) \
/ (X.max() - X.min())
j = lut.take( xscale.astype(int), mode="clip" ) # take(floats) in numpy 1.7 ?
#---------------------------------------------------------------------------
# X | | | | |
# j 0 1 2 3 4 ...
# LUT |....|.......|.|.............|.... -> int j (+ offset in [0, 1) )
#---------------------------------------------------------------------------
searchsorted非常快，时间~2,500，
所以这可能不会快得多。
但是LUT在C中非常非常，这是一种简单的速度/内存权衡。