python - 在Python中使用插值统一ct扫描体素大小

Question

我在Matlab中使用了interp2，例如以下代码，这是@rayryeng的答案的一部分: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 大小语法:(行、列、第三维切片)和 Python 大小语法:(第三维切片、行、列)。但是，将Matlab代码转换为Python代码后出现错误，ValueError: Invalid length for input z for nonquartz 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

出了什么问题？非常感谢。

Answer 1

在 MATLAB 中，interp2有作为参数:

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

您仅使用后 3 个参数，前两个参数假定为 input_x = 1:size(input_z,2) 和 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，并返回一个对象，您可以调用该对象来获取插值:

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

按照文档中的示例，我得到如下结果:

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，如有错误请告知。]