插值三维体积与numpy的和或SciPy的 [英] interpolate 3D volume with numpy and or scipy

问题描述

我感到非常沮丧，因为几个小时后，我似乎无法能够做一个看似简单的3D插值蟒蛇。在Matlab的一切我所要做的就是

I am extremely frustrated because after several hours I can't seem to be able to do a seemingly easy 3D interpolation in python. In Matlab all I had to do was

Vi = interp3(x,y,z,V,xi,yi,zi)

这样做有什么用SciPy的的ndimage.map_coordinate或其他numpy的方法完全等效？

What is the exact equivalent of this using scipy's ndimage.map_coordinate or other numpy methods?

感谢

推荐答案

在SciPy的0.14或更高版本，有一个新的功能，<一个href="http://docs.scipy.org/doc/scipy-0.14.0/reference/generated/scipy.interpolate.RegularGridInterpolator.html#scipy.interpolate.RegularGridInterpolator"相对=nofollow> scipy.interpolate.RegularGridInterpolator 这非常类似于 interp3 。

In scipy 0.14 or later, there is a new function scipy.interpolate.RegularGridInterpolator which closely resembles interp3.

MATLAB命令 VI = interp3（X，Y，Z，V，十一，苡仁，紫）将转化是这样的：

The MATLAB command Vi = interp3(x,y,z,V,xi,yi,zi) would translate to something like:

from numpy import array
from scipy.interpolate import RectangularGridInterpolator as rgi
my_interpolating_function = rgi((x,y,z), V)
Vi = my_interpolating_function(array([xi,yi,zi]).T)

下面是一个完整的例子来展示这两种;它会帮助你了解确切的差异...

Here is a full example demonstrating both; it will help you understand the exact differences...

MATLAB code：

MATLAB CODE:

x = linspace(1,4,11);
y = linspace(4,7,22);
z = linspace(7,9,33);
V = zeros(22,11,33);
for i=1:11
    for j=1:22
        for k=1:33
            V(j,i,k) = 100*x(i) + 10*y(j) + z(k);
        end
    end
end
xq = [2,3];
yq = [6,5];
zq = [8,7];
Vi = interp3(x,y,z,V,xq,yq,zq);

结果是 VI = [268 357] 这确实是在这两点（2,6,8）的值和（3,5,7）。

The result is Vi=[268 357] which is indeed the value at those two points (2,6,8) and (3,5,7).

MATPLOTLIB code：

MATPLOTLIB CODE:

from scipy.interpolate import RegularGridInterpolator
from numpy import linspace, zeros, array
x = linspace(1,4,11)
y = linspace(4,7,22)
z = linspace(7,9,33)
V = zeros((11,22,33))
for i in range(11):
    for j in range(22):
        for k in range(33):
            V[i,j,k] = 100*x[i] + 10*y[j] + z[k]
fn = RegularGridInterpolator((x,y,z), V)
pts = array([[2,6,8],[3,5,7]])
print(fn(pts))

同样是 [268357] 。所以你看到一些细微的差别：Matplotlib使用X，Y，Z指数为了在MATLAB用Y，X，Z（奇怪）;在Matplotlib定义在一个单独的步骤中的函数，当你调用它，坐标分组像（X1，Y1，Z1），（X2，Y2，Z2），...而MATLAB的使用（X1，X2，... ），（Y1，Y2，...），（Z1，Z2，...）。

Again it's [268,357]. So you see some slight differences: Matplotlib uses x,y,z index order while MATLAB uses y,x,z (strangely); In Matplotlib you define a function in a separate step and when you call it, the coordinates are grouped like (x1,y1,z1),(x2,y2,z2),... while matlab uses (x1,x2,...),(y1,y2,...),(z1,z2,...).

除此之外，两者是相似的，并同样易于使用。

Other than that, the two are similar and equally easy to use.

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