matrix using python

Question

I'm new to python and I'm writing a program fro matrix but there is a problem I don't know to get the right output and I need help with it. this is the question:Given a nXn matrix A and a kXn matrix B find AB . and here is what I have so far. Thank you in advance

def matrixmult (A, B):
    rows_A = len(A)
    cols_A = len(A[0])
    rows_B = len(B)
    cols_B = len(B[0])

    if cols_A != rows_B:
      print "Cannot multiply the two matrices. Incorrect dimensions."
      return

    # Create the result matrix
    # Dimensions would be rows_A x cols_B
    C = [[0 for row in range(cols_B)] for col in range(rows_A)]
    print C

    for i in range(rows_A):
        for j in range(cols_B):
            for k in range(cols_A):
                C[i][j] += A[i][k]*B[k][j]
    return C

I don't see anything obviously wrong here. Do you have an example input where it gets the wrong answer? — Danica
– Danica, Commented Apr 1, 2013 at 2:18
When you say you get no output, do you mean it doesn't print anything, that the return value is None, or something else? Oh, and in the code you posted, return C should be on column 4 instead of 3. Typo in your post? — kelm
– kelm, Commented Apr 1, 2013 at 2:28

Community · Accepted Answer · 2017-05-23 12:31:24Z

1

Your function:

def matrixmult (A, B):
    rows_A = len(A)
    cols_A = len(A[0])
    rows_B = len(B)
    cols_B = len(B[0])

    if cols_A != rows_B:
      print "Cannot multiply the two matrices. Incorrect dimensions."
      return

    # Create the result matrix
    # Dimensions would be rows_A x cols_B
    C = [[0 for row in range(cols_B)] for col in range(rows_A)]
    print C

    for i in range(rows_A):
        for j in range(cols_B):
            for k in range(cols_A):
                C[i][j] += A[i][k]*B[k][j]
    return C

Which appears to be the same as this function.

If I run this:

matrix=[[1,2,3],
    [4,5,6],
    [7,8,9]]

print matrixmult(matrix, matrix)    # that is your function...

It returns:

[[30, 36, 42], [66, 81, 96], [102, 126, 150]]

This is the same as Numpy:

import numpy as np

a=np.array(matrix)
b=np.array(matrix)
print np.dot(a,b)
#  [[ 30  36  42]
    [ 66  81  96]
    [102 126 150]]

And the same as the matrix multiply more tersely stated:

def mult(mtx_a,mtx_b):
    tpos_b = zip( *mtx_b)
    rtn = [[ sum( ea*eb for ea,eb in zip(a,b)) for b in tpos_b] for a in mtx_a]
    return rtn

So -- it is probably your input data that is the issue.

edited May 23, 2017 at 12:31

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answered Apr 1, 2013 at 4:28

dawg

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1 Comment

Mido Over a year ago

yeah that was my issue, you're really explained it well Thanks!

Carlos Neves · Accepted Answer · 2013-04-01 02:33:03Z

0

Use numPy library to solve your problem.

import numpy as np

x = np.array( ((2,3), (3, 5)) )

y = np.array( ((1,2), (5, -1)) )

print x * y

array([[ 2, 6], [15, -5]])

More examples: http://www.python-course.eu/matrix_arithmetic.php

Download numPy: http://scipy.org/Download

answered Apr 1, 2013 at 2:33

Carlos Neves

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2 Comments

DSM Over a year ago

While numpy is definitely the way to go in production code, I'm pretty sure this is an exercise, so going right to a library might defeat the purpose. Moreover, the OP is doing matrix multiplication, not elementwise multiplication, so x*y on two ndarrays wouldn't work. You'd need to use dot or the matrix class.

Carlos Neves Over a year ago

Mido you need to go to numpy website and download the numpy library. Afther you need install and now you can go to your project and do the import.

Community · Accepted Answer · 2020-06-20 09:12:55Z

0

One liner:

def matrixmult(m1, m2):
    return [
        [sum(x * y for x, y in zip(m1_r, m2_c)) for m2_c in zip(*m2)] for m1_r in m1
    ]

Explanation:

zip(*m2) - gets a column from the second matrix

zip(m1_r, m2_c) - creates tuple from m1 row and m2 column

sum(...) - sums multiplication row * col

Test:

m1 = [[1, 2, 3], [4, 5, 6]]
m2 = [[7, 8], [9, 10], [11, 12]]
result = matrixmult(m1, m2)
assert result == [[58, 64], [139, 154]]

edited Jun 20, 2020 at 9:12

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answered Jan 26, 2019 at 15:18

Vlad Bezden

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