Hi,

I am trying to extract values from a matrix contained at the position
dictated in a vector.
So if I have a matrix and vector with values:
mymatrix = [3 4 5 6 7; 2 3 4 5 6]
myvector = [2; 4]

I want a resulting vector:
result = [4; 5]

I can do it using a for loop:
for i=1:2;
  result(i)=mymatrix(i,myvector(i));
end;

Is there a more elegant and efficient way of doing this?

Cyrille MdC

In article <fba4908c-d446-4be4-a2d1-4df099affa53@g7g2000yqa.googlegroups.com>,
Cyrille  <cyrillem@gmail.com> wrote:
>Hi,
>
>I am trying to extract values from a matrix contained at the position
>dictated in a vector.
>So if I have a matrix and vector with values:
>mymatrix = [3 4 5 6 7; 2 3 4 5 6]
>myvector = [2; 4]
>
>I want a resulting vector:
>result = [4; 5]
>
>I can do it using a for loop:
>for i=1:2;
>  result(i)=mymatrix(i,myvector(i));
>end;
>
>Is there a more elegant and efficient way of doing this?
>
>Cyrille MdC

It took me a minute to see what wanted. Look at
newmatrix = mymatrix'
newvector = myvector + 5*(0:1)'
newmatrix(newvector)

5 is the of rows in newmatrix or columns in mymatrix.
1 is one less then the number of entries in myvector.
-- 
Steven Bellenot                 http://www.math.fsu.edu/~bellenot
Professor and Associate Chair               phone: (850) 644-7405 
Department of Mathematics                        office: 223 Love
Florida State University          email: bellenot at math.fsu.edu

On Fri, 11 Dec 2009 08:37:21 -0800, Cyrille wrote:

> Hi,
> 
> I am trying to extract values from a matrix contained at the position
> dictated in a vector.
> So if I have a matrix and vector with values: mymatrix = [3 4 5 6 7; 2 3
> 4 5 6]
> myvector = [2; 4]
> 
> I want a resulting vector:
> result = [4; 5]
> 
> I can do it using a for loop:
> for i=1:2;
>   result(i)=mymatrix(i,myvector(i));
> end;
> 
> Is there a more elegant and efficient way of doing this?
> 
> Cyrille MdC

It would seem that your myvector, having only two entries, would only 
specify one location in the matrix: mymatrix(myvector(1), myvector(2)).

If you specified 

myvector = [[1 2]' [5 1]'];
then

mymatrix(myvector(:, 1), myvector(:, 2))

would cough up mymatrix(1, 5) and mymatrix(2, 1).



-- 
www.wescottdesign.com

On Dec 11, 6:11=A0pm, belle...@math.fsu.edu (Steve Bellenot) wrote:
> In article <fba4908c-d446-4be4-a2d1-4df099aff...@g7g2000yqa.googlegroups.=
com>,
>
>
>
> Cyrille =A0<cyril...@gmail.com> wrote:
> >Hi,
>
> >I am trying to extract values from a matrix contained at the position
> >dictated in a vector.
> >So if I have a matrix and vector with values:
> >mymatrix =3D [3 4 5 6 7; 2 3 4 5 6]
> >myvector =3D [2; 4]
>
> >I want a resulting vector:
> >result =3D [4; 5]
>
> >I can do it using a for loop:
> >for i=3D1:2;
> > =A0result(i)=3Dmymatrix(i,myvector(i));
> >end;
>
> >Is there a more elegant and efficient way of doing this?
>
> >Cyrille MdC
>
> It took me a minute to see what wanted. Look at
> newmatrix =3D mymatrix'
> newvector =3D myvector + 5*(0:1)'
> newmatrix(newvector)
>
> 5 is the of rows in newmatrix or columns in mymatrix.
> 1 is one less then the number of entries in myvector.
> --
> Steven Bellenot =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0http://www.math.fsu.edu/~b=
ellenot
> Professor and Associate Chair =A0 =A0 =A0 =A0 =A0 =A0 =A0 phone: (850) 64=
4-7405
> Department of Mathematics =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0=
office: 223 Love
> Florida State University =A0 =A0 =A0 =A0 =A0email: bellenot at math.fsu.e=
du

Thank you Steven,

That works - it definitely took me a few minutes to understand the
strategy you outlined.
I really enjoyed your idea about multiplying the incremental vector
(0:2) by the transposed matrix' row number. I didn't realize you could
access values by wrapping through the matrix like this. Understanding
your algorithm gave me a buzz - thank you again.
I'm applying it to a 6000x30 matrix. Although I originally thought
your strategy might not be as fast as the for loop option, I timed
both methods and yours takes 63ms to my 172ms. Almost a third the
time!

Have a great day and weekend!
Cyrille

On Dec 11, 6:11=A0pm, belle...@math.fsu.edu (Steve Bellenot) wrote:
> In article <fba4908c-d446-4be4-a2d1-4df099aff...@g7g2000yqa.googlegroups.=
com>,
>
>
>
> Cyrille =A0<cyril...@gmail.com> wrote:
> >Hi,
>
> >I am trying to extract values from a matrix contained at the position
> >dictated in a vector.
> >So if I have a matrix and vector with values:
> >mymatrix =3D [3 4 5 6 7; 2 3 4 5 6]
> >myvector =3D [2; 4]
>
> >I want a resulting vector:
> >result =3D [4; 5]
>
> >I can do it using a for loop:
> >for i=3D1:2;
> > =A0result(i)=3Dmymatrix(i,myvector(i));
> >end;
>
> >Is there a more elegant and efficient way of doing this?
>
> >Cyrille MdC
>
> It took me a minute to see what wanted. Look at
> newmatrix =3D mymatrix'
> newvector =3D myvector + 5*(0:1)'
> newmatrix(newvector)
>
> 5 is the of rows in newmatrix or columns in mymatrix.
> 1 is one less then the number of entries in myvector.
> --
> Steven Bellenot =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0http://www.math.fsu.edu/~b=
ellenot
> Professor and Associate Chair =A0 =A0 =A0 =A0 =A0 =A0 =A0 phone: (850) 64=
4-7405
> Department of Mathematics =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0=
office: 223 Love
> Florida State University =A0 =A0 =A0 =A0 =A0email: bellenot at math.fsu.e=
du

Thank you Steven,

That works - it definitely took me a few minutes to understand the
strategy you outlined.
I really enjoyed your idea about multiplying the incremental vector
(0:2) by the transposed matrix' row number. I didn't realize you could
access values by wrapping through the matrix like this. Understanding
your algorithm gave me a buzz - thank you again.
I'm applying it to a 6000x30 matrix. Although I originally thought
your strategy might not be as fast as the for loop option, I timed
both methods and yours takes 63ms to my 172ms. Almost a third the
time!

Have a great day and weekend!
Cyrille