MATLAB Central Newsreader - adding a vector to each col of a previous vector

adding a vector to each col of a previous vector

Adrian — Wed, 04 Nov 2009 17:19:03 -0500

I use too many loops all the time and I want to become more familiar with vector notation and matrix manipulation.
So I would like to know the following.
Without using a loop what is the most elegant (and efficient) way to add a new vector to each col of another vector.
For example:
a = [1 5 20 35 100]
b = 0:4;

c = [1 5 20 35 100
       2 6 21 36 101
       3 7 22 37 102
       4 8 23 38 103
       5 9 24 39 104]

This is probably pretty easy. Help me out if you can.

Re: adding a vector to each col of a previous vector

Matt — Wed, 04 Nov 2009 17:31:04 -0500

"Adrian " <adrian.suszko@uhn.on.ca> wrote in message <hcsd27$15e$1@fred.mathworks.com>...
> I use too many loops all the time and I want to become more familiar with vector notation and matrix manipulation.
> So I would like to know the following.
> Without using a loop what is the most elegant (and efficient) way to add a new vector to each col of another vector.
> For example:
> a = [1 5 20 35 100]
> b = 0:4;
>
> c = [1 5 20 35 100
> 2 6 21 36 101
> 3 7 22 37 102
> 4 8 23 38 103
> 5 9 24 39 104]
>
> This is probably pretty easy. Help me out if you can.

The easiest is bsxfun(@plus, b',a);

If even only one of a or b has small dimensions, however, for-loops have been found to be more efficient.

Re: adding a vector to each col of a previous vector

Nasser M. Abbasi — Wed, 04 Nov 2009 17:35:43 -0500

"Adrian " <adrian.suszko@uhn.on.ca> wrote in message
news:hcsd27$15e$1@fred.mathworks.com...
>I use too many loops all the time and I want to become more familiar with
>vector notation and matrix manipulation.
> So I would like to know the following.
> Without using a loop what is the most elegant (and efficient) way to add a
> new vector to each col of another vector.

The name vector is usually used for stuff that has one row or column. For
stuff with more than one of these, it is called a matrix, not a vector.
Altough a vector can also be considered a matrix, but normally we call a
matrix something that has more than one row or column in it. So 'c' here is
a matrix not a vector.

> For example:
> a = [1 5 20 35 100]
> b = 0:4;
>
> c = [1 5 20 35 100
> 2 6 21 36 101
> 3 7 22 37 102
> 4 8 23 38 103
> 5 9 24 39 104]
>
> This is probably pretty easy. Help me out if you can.

To add 'a' to bottom of c matrix, one way is to repmat it below c as follows

[nRow,nCol]=size(c);
newc=[c;repmat(a',[1 nCol])]

newc =
     1 5 20 35 100
     2 6 21 36 101
     3 7 22 37 102
     4 8 23 38 103
     5 9 24 39 104
     1 1 1 1 1
     5 5 5 5 5
    20 20 20 20 20
    35 35 35 35 35
   100 100 100 100 100

similary for b

--Nasser

Re: adding a vector to each col of a previous vector

Nathan — Wed, 04 Nov 2009 18:07:32 -0500

On Nov 4, 9:35 am, "Nasser M. Abbasi" <n...@12000.org> wrote:
> "Adrian " <adrian.sus...@uhn.on.ca> wrote in message
>
> news:hcsd27$15e$1@fred.mathworks.com...
>
> >I use too many loops all the time and I want to become more familiar with
> >vector notation and matrix manipulation.
> > So I would like to know the following.
> > Without using a loop what is the most elegant (and efficient) way to add a
> > new vector to each col of another vector.
>
> The name vector is usually used for stuff that has one row or column. For
> stuff with more than one of these, it is called a matrix, not a vector.
> Altough a vector can also be considered a matrix, but normally we call a
> matrix something that has more than one row or column in it. So 'c' here is
> a matrix not a vector.
>
> > For example:
> > a = [1 5 20 35 100]
> > b = 0:4;
>
> > c = [1 5 20 35 100
> > 2 6 21 36 101
> > 3 7 22 37 102
> > 4 8 23 38 103
> > 5 9 24 39 104]
>
> > This is probably pretty easy. Help me out if you can.
>
> To add 'a' to bottom of c matrix, one way is to repmat it below c as follows
>
> [nRow,nCol]=size(c);
> newc=[c;repmat(a',[1 nCol])]
>
> newc =
> 1 5 20 35 100
> 2 6 21 36 101
> 3 7 22 37 102
> 4 8 23 38 103
> 5 9 24 39 104
> 1 1 1 1 1
> 5 5 5 5 5
> 20 20 20 20 20
> 35 35 35 35 35
> 100 100 100 100 100
>
> similary for b
>
> --Nasser

I think he meant from a and b, he gets c
ex:
c = [a+b(1);a+b(2);a+b(3);a+b(4);a+b(5)]
%%%%%%%%%%%%%%%%%%%%%%%
c =
     1 5 20 35 100
     2 6 21 36 101
     3 7 22 37 102
     4 8 23 38 103
     5 9 24 39 104

I dunno. I think for loops work wonderfully for this.

initialize c:
c = zeros(length(a),length(b));
for i=1:length(b)
c(i,:) = a+b(i)
end

-Nathan

Re: adding a vector to each col of a previous vector

Adrian — Wed, 04 Nov 2009 18:10:19 -0500

"Nasser M. Abbasi" <nma@12000.org> wrote in message <QXiIm.7585$fE2.5316@newsfe04.iad>...
>
> "Adrian " <adrian.suszko@uhn.on.ca> wrote in message
> news:hcsd27$15e$1@fred.mathworks.com...
> >I use too many loops all the time and I want to become more familiar with
> >vector notation and matrix manipulation.
> > So I would like to know the following.
> > Without using a loop what is the most elegant (and efficient) way to add a
> > new vector to each col of another vector.
>
> The name vector is usually used for stuff that has one row or column. For
> stuff with more than one of these, it is called a matrix, not a vector.
> Altough a vector can also be considered a matrix, but normally we call a
> matrix something that has more than one row or column in it. So 'c' here is
> a matrix not a vector.
>
> > For example:
> > a = [1 5 20 35 100]
> > b = 0:4;
> >
> > c = [1 5 20 35 100
> > 2 6 21 36 101
> > 3 7 22 37 102
> > 4 8 23 38 103
> > 5 9 24 39 104]
> >
> > This is probably pretty easy. Help me out if you can.
>
> To add 'a' to bottom of c matrix, one way is to repmat it below c as follows
>
> [nRow,nCol]=size(c);
> newc=[c;repmat(a',[1 nCol])]
>
> newc =
> 1 5 20 35 100
> 2 6 21 36 101
> 3 7 22 37 102
> 4 8 23 38 103
> 5 9 24 39 104
> 1 1 1 1 1
> 5 5 5 5 5
> 20 20 20 20 20
> 35 35 35 35 35
> 100 100 100 100 100
>
> similary for b
>
> --Nasser
>
Apologies I rushed that post and didn't include the fact that I meant that I wanted my result to be a matrix which in this case was the "c" that i described above.