Replacing rows of a numpy array

Question

I'm trying to emulate Cramer's rule for indefinite number of variables. This is what I have so far

def cramers(n)
    matrix = []
    for i in range(0, n):
        matrix.append([])
        for j in range(0, n+1):
            matrix[i].append(0)
            matrix[i][j] = float(input("V" + str(i+1) + str(j+1) + ": "))
    #n = 2, V11 = 1, V12 = 2, V13 = 3, V21 = 4, V22 = 5, V23 = 6

    mat = np.array(matrix)
    matA = mat[0:n, 0:n]
    matB = mat[0:n, n:]

    matATrans = matA.transpose()
    matBTrans = matB.transpose()

for n = 2, these are sample input values of matATrans and matBTrans:

matATrans = [[1. 4.], [2. 5.]]
matBTrans = [3. 6.]

My question is, how can I produce a numpy array, mat, with n = 2 length for instance, whose value contains this?

mat = [[[3. 6.],[2. 5.]],[[1. 4.],[3. 6.]]]

Basically, the (n-1)th element of matATrans is replaced by matBTrans. I'm thinking it could be done using a forloop. This was my attempt.

for i in range(n):
    matATrans[i] = matBTrans
    print(matATrans)

The result is this:

[3. 6.],[2. 5.]
[3. 6.],[3. 6.]

Obviously, it's wrong because the content of matATrans has been altered in the first loop. Also, it still isn't appended to a mat matrix.

Please enlighten me. Thanks

I'm trying to program my own solver for system of linear equations using Cramer's rule — Joe Kerr
– Joe Kerr, Commented Mar 16, 2020 at 3:39

ferhen · Accepted Answer · 2020-03-16 04:31:16Z

1

In python, when you assign a list to a new variable you're just creating another reference to the same object, that is the reason for the current behavior. One way to fix this is to use the copy method and copy the matrix A each loop.

n = 2
matATrans = [[1, 4], [2, 5]]
matBTrans = [3, 6]

result = []
for i in range(n):
    temp = matATrans.copy()
    temp[i] = matBTrans
    result.append(temp)

print(result)

edited Mar 16, 2020 at 4:31

answered Mar 16, 2020 at 4:23

ferhen

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