如何将以下Matlab循环转换为Python？

Question

I have been working on translating Matlab code into Python and came across a loop that I'm having some difficulty converting as I'm fairly new to both the languages.

if fdp >=2
degreeTwoVector=[];
counter =1;
for i = 1:numVariables
    for j = 1:numVariables
        degreeTwoVector(counter,:) = [i j 0];
        counter = counter +1;
    end
end

sortedDegreeTwoVector = sort(degreeTwoVector,2);
degreeTwoVector = unique(sortedDegreeTwoVector, 'rows');

combinationVector = [combinationVector; degreeTwoVector];
end

Here's what I could come up with while converting it to python(incomplete):

if fdp >= 2:
    degreeTwoVector = np.array([])
    counter = 1
    for i in range(1, numVariables+1):
        for j in range(1, numVariables+1):
        degreeTwoVector(counter, :) = np.array([i, j, 0])
        counter = counter + 1
        break
    sortedDegreeTwoVector = degreeTwoVector[np.argsort(degreeTwoVector[:, 1])]

I certainly know there are some mistakes in it. So I'd be grateful if you could help me complete the conversion and correct any mistakes. Thanks in advance!

Answer 1

You are not too far off: You do not need a break statement, it is causing a precocious, well, break to the loop (at the first iteration). So here you go:

numVariables = np.shape(X)[0] #  number of rows in X which is given
if fdp >= 2:
    degreeTwoVector = np.zeros((numVariables, 3)) #  you need to initialize the shape
    counter = 0 # first index is 0
    for i in range(numVariables):
        for j in range(numVariables):
            degreeTwoVector[counter, :] = np.array([i, j, 0])
            counter = counter + 1 #  counter += 1 is more pythonic
    sortedDegreeTwoVector = np.sort(degreeTwoVector, axis=1);
    degreeTwoVector = np.vstack({tuple(row) for row in sortedDegreeTwoVector})

    combinationVector = np.vstack((combinationVector, degreeTwoVector))

EDIT : added equivalent of code outside the loop in the original question.

Apart from the fact that i don't see where you defined combinationVector , everything should be okay.