我想将以下 MATLAB 代码转换为 python？ 这是正确的方法吗？

Question

How can I change this part [AIF,j]=get_AIF_j(InterpFact) and [~,j_index] = min(InterpFact-AIF_vect) correctly? And what about the remaining code? Thanks in advance.

%Matlab code
InterpFact = (fs_h/2/2)/(fd_max); 
[AIF,j]=get_AIF_j(InterpFact);    
    
function [AIF,j] = get_AIF_j (InterpFact)
j_vect = 1:10;
AIF_vect = floor(j_vect*InterpFact)./j_vect;
[~,j_index] = min(InterpFact-AIF_vect);
j = j_vect(j_index);
AIF = AIF_vect(j_index);
end

#Python code
InterpFact = (fs_h/2/2)/(fd_max) 
[AIF,j]=get_AIF_j(InterpFact)
    
def get_AIF_j (InterpFact):
    j_vect =np.arange(1,11)
    AIF_vect = np.floor(j_vect*InterpFact)/j_vect
    [~,j_index] = min(InterpFact-AIF_vect)
    j = j_vect[j_index]
    AIF = AIF_vect[j_index];
    return AIF,j

Answer 1

Try this to see if it delivers what it should (I am not sure here as I am not fluent in matlab):

#Python code
import numpy as np
def get_AIF_j (InterpFact):
    j_vect   = np.arange(1,11)
    AIF_vect = np.floor(j_vect*InterpFact)/j_vect
    j_index = int( min(InterpFact-AIF_vect) )
    print(j_index)
    j = j_vect[j_index]
    AIF = AIF_vect[j_index];
    return AIF, j

fs_h = 24; fd_max = 1
InterpFact = (fs_h/2/2)/(fd_max) 
AIF, j = get_AIF_j(InterpFact)
print(AIF,j)    

gives:

0
6.0 1

Answer 2

This MATLAB:

[~,j_index] = min(InterpFact-AIF_vect);

would be translated to Python as:

j_index = np.argmin(InterpFact-AIF_vect)

Also, …/(fd_max) can only be translated the way you did if fd_max is a scalar. A division with a matrix in MATLAB solves a system of linear equations.

I strongly recommend that you run the two pieces of code side by side with the same input, to verify that they do the same thing. You cannot go by guesses as to what a piece of code does.