12

我有一个 3D 数组，我需要在一个轴（最后一个维度）上进行插值。比方说y.shape = (nx, ny, nz)，我想nz对每个进行插值(nx, ny)。但是，我想在每个[i, j].

这是一些示例代码。如果我想插入单个值，比如说new_z，我会这样scipy.interpolate.interp1d使用

# 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)

但是，对于这个问题，我真正想要的是new_z为每个y[i, j]. 所以我这样做：

# y is a 3D ndarray
# x is a 1D ndarray with the abcissa values
# new_z is a 2D array
result = numpy.empty(y.shape[:-1])
for i in range(nx):
    for j in range(ny):
        f = scipy.interpolate.interp1d(x, y[i, j], axis=-1, kind='linear')
        result[i, j] = f(new_z[i, j])

不幸的是，对于多个循环，这变得低效且缓慢。有没有更好的方法来做这种插值？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并且不想在 Cython 中手动完成。

4

6 回答 6

8

为了加快高阶插值，您可以interp1d()只调用一次，然后使用模块中的_spline属性和低级函数。这是代码：_bspleval()_fitpack

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
于 2012-11-22T01:41:33.277 回答
4

我不认为interp1d有一种方法可以快速做到这一点，所以你不能避免这里的循环。

您可能仍然可以通过使用 编码线性插值来避免 Cython np.searchsorted，如下所示（未测试）：

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，对三次插值没有帮助。

编辑：使它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

不过，有人可能会争辩说 Cython 代码更容易阅读。

于 2012-11-21T13:10:20.433 回答
3

由于上面的 numpy 建议花费了太长时间，我可以等待，所以这里是 cython 版本以供将来参考。从一些松散的基准来看，它的速度大约快了 3000 倍（当然，它只是线性插值，并没有那么多，interp1d但可以用于此目的）。

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
于 2012-11-21T13:51:28.620 回答
3

以@pv. 的回答为基础，并对内部循环进行矢量化，以下给出了显着的加速（编辑：将昂贵的更改numpy.tile为 using 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()))

这在我的机器上给出了以下时间：

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

转到和nx = 300，可以得到 130 倍的加速：ny = 300nz = 500

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

您需要编写自己的三次插值算法，但这不应该那么难。

于 2012-11-21T14:37:21.967 回答
2

尽管有几个不错的答案，但他们仍然在一个固定的 500 长数组中进行 250k 插值：

j250k = np.searchsorted( X500, X250k )  # indices in [0, 500)

这可以通过 LUT、查找表（例如 5k 个插槽）来加速：

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相当快，时间〜ln2 500，所以这可能不会快多少。
但是 LUT 在 C 中非常快，这是一个简单的速度/内存权衡。

于 2013-02-07T17:43:59.813 回答
2

您可以为此使用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.)
于 2012-11-21T13:52:49.293 回答