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Division matrix

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  Division matrix Cleuson Alves <cleuson.o@gmail.com> - 2012-11-12 17:00 -0800
    Re: Division matrix Ian Kelly <ian.g.kelly@gmail.com> - 2012-11-12 18:25 -0700
    Re: Division matrix Dennis Lee Bieber <wlfraed@ix.netcom.com> - 2012-11-13 13:48 -0500
    Re: Division matrix "R. Michael Weylandt" <michael.weylandt@gmail.com> - 2012-11-13 22:14 +0000
    Re: Division matrix wxjmfauth@gmail.com - 2012-11-14 01:04 -0800

#33209 — Division matrix

Hello, I need to solve an exercise follows, first calculate the inverse matrix and then multiply the first matrix.
I await help.
Thank you.
follows the code below incomplete.

m = [[1,2,3],[4,5,6],[7,8,9]]
x = []
for i in [0,1,2]:
    y = []
    for linha in m:
        y.append(linha[i])
    x.append(y)

print x
[[1, 4, 7], [2, 5, 8], [3, 6, 9]]

def ProdMatrix(x,b):
    tamL = len(x)
    tamC = len(x[0])
    c = nullMatrix(tamL,tamC)
    for i in range(tamL):
        for j in range(tamC):
            val = 0
            for k in range(len(b)):
                val = val + x[i][l]*b[k][j]
            c[i][j]
    return c

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#33210

On Mon, Nov 12, 2012 at 6:00 PM, Cleuson Alves <cleuson.o@gmail.com> wrote:
> Hello, I need to solve an exercise follows, first calculate the inverse matrix and then multiply the first matrix.
> I await help.
> Thank you.
> follows the code below incomplete.

So what is the specific problem with the code that you're looking for help with?

> m = [[1,2,3],[4,5,6],[7,8,9]]
> x = []
> for i in [0,1,2]:
>     y = []
>     for linha in m:
>         y.append(linha[i])
>     x.append(y)
>
> print x
> [[1, 4, 7], [2, 5, 8], [3, 6, 9]]

This calculates the transpose of the matrix, not the inverse.  If the
inverse is really what you're after, you should start by reading up on
the math to compute that.

Note that "for i in [0,1,2]:" could be more generally written as "for
i in range(len(m[0])):", and then you don't need to rewrite your for
loop if the dimensions of m change.

If you actually do want the transpose, then I'll also mention in
passing that there is a very nice one-liner using zip() to do this.
I'll leave the details as an exercise. ;-)

> def ProdMatrix(x,b):
>     tamL = len(x)
>     tamC = len(x[0])
>     c = nullMatrix(tamL,tamC)
>     for i in range(tamL):
>         for j in range(tamC):
>             val = 0
>             for k in range(len(b)):
>                 val = val + x[i][l]*b[k][j]
>             c[i][j]
>     return c

In the multiplication line you're using the variable 'l' as an index,
but you haven't defined it.  Probably you wanted 'k' here.

You're also discarding 'val' after adding it up.  If you want that
value in the 'c' matrix, then you need to store it there before going
on to the next loop.

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#33258

On Tue, 13 Nov 2012 10:19:43 -0200, Cleuson Alves <cleuson.o@gmail.com>
declaimed the following in gmane.comp.python.general:

> Thanks, I'm starting to plan now, so I'm still confused with the production
> code, but what I need is to divide array 2x2 or 3x3.
> I still can not!

	Divide it by what? A scalar... Another square matrix of the same
size... Or a square matrix of a different size (is that even
possible?)...

	Based
http://en.wikipedia.org/wiki/Division_%28mathematics%29#Division_of_matrices
 upon, I can understand where the need for the inverse comes from -- and
multiplication by the inverse gives the "division". Next up,
http://en.wikipedia.org/wiki/Invertible_matrix#Inversion_of_2.C3.972_matrices
gives direct formulations for 2x2 and 3x3 matrices.

-- 
	Wulfraed                 Dennis Lee Bieber         AF6VN
        wlfraed@ix.netcom.com    HTTP://wlfraed.home.netcom.com/

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#33269

On Tue, Nov 13, 2012 at 1:00 AM, Cleuson Alves <cleuson.o@gmail.com> wrote:
> Hello, I need to solve an exercise follows, first calculate the inverse matrix and then multiply the first matrix.

I would just point out that in most numerical applications, you rarely
need to calculate the intermediate of the matrix inverse directly.
See, e.g., http://www.johndcook.com/blog/2010/01/19/dont-invert-that-matrix/

Of course, if this hasn't been said yet: NumPy.

Michael

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#33316

Le mardi 13 novembre 2012 02:00:28 UTC+1, Cleuson Alves a écrit :
> Hello, I need to solve an exercise follows, first calculate the inverse matrix and then multiply the first matrix.
> 
> I await help.
> 
> Thank you.
> 
> follows the code below incomplete.
> 
> 
> 
> m = [[1,2,3],[4,5,6],[7,8,9]]
> 
> x = []
> 
> for i in [0,1,2]:
> 
>     y = []
> 
>     for linha in m:
> 
>         y.append(linha[i])
> 
>     x.append(y)
> 
> 
> 
> print x
> 
> [[1, 4, 7], [2, 5, 8], [3, 6, 9]]
> 
> 
> 
> def ProdMatrix(x,b):
> 
>     tamL = len(x)
> 
>     tamC = len(x[0])
> 
>     c = nullMatrix(tamL,tamC)
> 
>     for i in range(tamL):
> 
>         for j in range(tamC):
> 
>             val = 0
> 
>             for k in range(len(b)):
> 
>                 val = val + x[i][l]*b[k][j]
> 
>             c[i][j]
> 
>     return c

------

Pedagogical hint:
Before blindly calculating the inverse matrix, it may be
a good idea to know if the inverse matrix exists.

jmf

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