def c(matrix):
    result = numpy.zeros(matrix.shape)

    sums = numpy.zeros(matrix.shape[0])

    for i in range(matrix.shape[0]):
        for j in range(matrix.shape[1]):
            sums[i] += matrix[i, j]

    for i in range(matrix.shape[0]):
        for j in range(matrix.shape[1]):
            result[i, j] = matrix[i, j] / sums[i]

    return result

Goal of the code

• The function c takes a 2D array called matrix.
• It returns a new 2D array (result) in which each row of matrix has been normalised to that its elements sum to 1.
• This means, for each row i, $\sum_j\text{result}[i,j]=1$.

Create a Result Array

result = numpy.zeros(matrix.shape)

• matrix.shape gives a tuple $(\text{num\_rows},\text{num\_cols})$.
• numpy.zeros(matrix, shape) initialises a 2D array (result) filled with zeros of the same dimensions as matrix.
• This ensures result is ready to store the normalised values.

Prepare an Array for the Row Sums

sums = numpy.zeros(matrix.shape[0])

• matrix.shape[0] is the number of rows in matrix.
• sums is a 1D array of length matrix.shape[0], initially all zeros.
• This will track the sum of each row.

Compute the Row Sums

for i in range(matrix.shape[0]):
    for j in range(matrix.shape[1]):
        sums[i] += matrix[i, j]