20190524 matrix algorithm, matrix addition, matrix multiplication, matrix transposition, etc

Keywords: Python

1. Transposition of two-dimensional matrix

arrA = [[1,2,3,4],[5,6,7,8],[9,10,11,12],[13,14,15,16]]
def turn(arr):
    if not arr:
        return []
    result = []
    for i in range(len(arr[0])):#Original column becomes row
        temp =[]
        for j in range(len(arr)):#Original row becomes column
    return result

2. Matrix addition, matrix A and matrix B need to be a matrix of N*M, that is, the rows and columns of the addition matrix must be equal

def matrix_add(arrA,arrB):
    if not arrA and not arrB:
        return []
    if len(arrA)!=len(arrB)or len(arrA[0])!=len(arrA[0]):
        return 'Error'
    arrC = [[None]*len(arrA[0]) for row in range(len(arrA))]#First define the result matrix
    for i in range(len(arrA)):
        for j in range(len(arrA[i])):
            arrC[i][j] = arrA[i][j] + arrB[i][j]
    return arrC
A = [[1,3,5,4],[7,9,1,3],[13,15,17,42]]
B = [[9,8,7,1],[6,5,4,2],[3,2,1,3]]

3. Matrix multiplication, matrix A and matrix B need to satisfy the condition that matrix A is m*n, matrix B is n*p, and result C is m*p

C11 = A11*B11+A12*B21+....+A1n*Bn1
C1P = A11*B1p+A12*B2p+...+A1n*Bnp
CMP = Am1*B1p+Am2*B2p+...+Amn*Bnp
The first index of arrA is equal to the first index of C, and the second index of arrA increases gradually each time
The first index of arrB increases gradually each time, and the second index of arrB is equal to the second index of C. So, because C is a matrix of m*p
First index of arrA = I
The second index of arrA = k
First index of arrB = k
The second index of arrB = J
A = [[1,3,5],[7,9,11],[13,15,17]]
B = [[9,8],[6,5],[3,2]]

def MatrixMultiply(arrA,arrB):
    if len(arrA[0])!=len(arrB):
        return False
    M = len(A)
    N = len(A[0])
    P = len(B[0])
    arrC = [[None] * P for row in range(M)]
    for i in range(len(arrA)):
        for j in range(len(arrB[0])):
            temp = 0
            for k in range(len(arrB)):
                #print(arrA[i][k],arrB[k][j],end =' ')
                temp = temp+int(arrA[i][k])*int(arrB[k][j])#Realization C1P = A11*B1p+A12*B2p+...+A1n*Bnp
            arrC[i][j] = temp
    return arrC


4. Write function and compress sparse matrix with trinomial
Sparse matrix: if most elements of a matrix are 0, it is a sparse matrix
Trinomial: nonzero terms are represented by (i, j, item value). If a sparse matrix has n nonzero terms, A(0:N,1:3) two-dimensional array can be used to store these nonzero terms
A(0,1) stores the number of rows of sparse matrix
A(0,2) stores the number of columns of sparse matrix
A(0,3) stores the nonzero term of sparse matrix
Each non-zero term is represented by (I, J, item value)

def Sparse_Transfer2_Trinomial(sparse):
    trinomial = []
    if not sparse:
        return trinomial
    non_zero = 0
    for i in range(len(sparse)):
        for j in range(len(sparse[i])):
            if sparse[i][j]:#sparse[i][j]Non 0
    return trinomial
Sparse = [[15,0,0,22,0,-15],[0,11,3,0,0,0],[0,0,0,-6,0,0],[0,0,0,0,0,0,0],[91,0,0,0,0,0],[0,0,28,0,0,0]]

5. Using trinomial transpose sparse matrix

Define the sparse matrix first, exchange the rows and columns, and fill the other positions with 0

def Turn_Sparse(trinomial):
    sparse = [[0]*trinomial[0][1] for i in range(trinomial[0][0])]
    for each in trinomial[1:]:
        sparse[each[1]][each[0]] = each[2]
    return sparse

Posted by jameslynns on Tue, 05 Nov 2019 13:58:35 -0800