通过在Python中使用插值来统一ct扫描体素大小 [英] Unification ct scan voxel size by using interpolation in Python

问题描述

我在Matlab中使用了interp2，例如以下代码，它是@rayryeng的答案的一部分:

I have used interp2 in Matlab, such as the following code, that is part of @rayryeng's answer in: Three dimensional (3D) matrix interpolation in Matlab:

d = size(volume_image)
[X,Y] = meshgrid(1:1/scaleCoeff(2):d(2), 1:1/scaleCoeff(1):d(1));
for ind = z
    %Interpolate each slice via interp2   
    M2D(:,:,ind) = interp2(volume_image(:,:,ind), X, Y);   
end

尺寸示例:

The image size is 512x512 and the number of slices is 133. So:
volume_image(rows, columns, slices in 3D dimenson) : 512x512x133 in 3D dimenson
X: 288x288
Y: 288x288
scaleCoeff(2): 0.5625
scaleCoeff(1): 0.5625
z = 1 up to 133 ,hence z: 1x133
ind: 1 up to 133
M2D(:,:,ind) finally is 288x288x133 in 3D dimenson

Aslo，Matlabs的语法为:(行，列，切片在第3维中)，Python语法的大小为:(切片在第3个暗部，行，列中). 但是，将Matlab代码转换为Python代码后，出现错误ValueError: Invalid length for input z for non rectangular grid:

Aslo, Matlabs syntax for size: (rows, columns, slices in 3rd dimenson) and Python syntax for size: (slices in 3rd dim, rows, columns). However, after convert the Matlab code to Python code occurred an error, ValueError: Invalid length for input z for non rectangular grid:

for ind in range(0, len(z)+1):
    M2D[ind, :, :] = interpolate.interp2d(X, Y, volume_image[ind, :, :]) # ValueError: Invalid length for input z for non rectangular grid

怎么了?非常感谢.

推荐答案

在MATLAB中，

In MATLAB, interp2 has as arguments:

result = interp2(input_x, input_y, input_z, output_x, output_y)

您仅使用后三个参数，假定前两个参数为input_x = 1:size(input_z,2)和input_y = 1:size(input_z,1).

You are using only the latter 3 arguments, the first two are assumed to be input_x = 1:size(input_z,2) and input_y = 1:size(input_z,1).

在Python中， scipy.interpolate.interp2 完全不同:它采用MATLAB函数的前3个输入参数，并返回一个您可以调用以获取内插值的对象:

In Python, scipy.interpolate.interp2 is quite different: it takes the first 3 input arguments of the MATLAB function, and returns an object that you can call to get interpolated values:

f = scipy.interpolate.interp2(input_x, input_y, input_z)
result = f(output_x, output_y)

按照文档中的示例，我得到如下信息:

Following the example from the documentation, I get to something like this:

from scipy import interpolate
x = np.arange(0, volume_image.shape[2])
y = np.arange(0, volume_image.shape[1])
f = interpolate.interp2d(x, y, volume_image[ind, :, :])
xnew = np.arange(0, volume_image.shape[2], 1/scaleCoeff[0])
ynew = np.arange(0, volume_image.shape[1], 1/scaleCoeff[1])
M2D[ind, :, :] = f(xnew, ynew)

[代码未经测试，如果有错误，请告诉我.]

[Code not tested, please let me know if there are errors.]

这篇关于通过在Python中使用插值来统一ct扫描体素大小的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！