Matrix with alternating signs in each row vector

Hi Guys,

Is there a way to improve on this code that I wrote to optimize it?

M = zeros(M,N); % create an MxN matrix 
M(1,:) = 1;  % Set first row to 1
for r = 2:I
    M(r,:) = -M(r-1,:);   %sets alternate rows to -1 and +1
end

a = M * diag(1 2 3 4 5);

so M creates:

M =

     1     1     1     1     1
    -1    -1    -1    -1    -1
     1     1     1     1     1
    -1    -1    -1    -1    -1
     1     1     1     1     1
    -1    -1    -1    -1    -1
     1     1     1     1     1
    -1    -1    -1    -1    -1

and a

a =

     1     2     3     4     5
    -1    -2    -3    -4    -5
     1     2     3     4     5
    -1    -2    -3    -4    -5
     1     2     3     4     5
    -1    -2    -3    -4    -5
     1     2     3     4     5
    -1    -2    -3    -4    -5

Is this the fastest and most efficient implementation to get the above? Thanks!

Answer 1

[Fangjun Jiang]

Some improvement.

m=5;n=4;
M=ones(m,n);
M(2:2:end,:)=-1

Or alternative:

m=9;n=8;
a=(2*mod((1:m)',2)-1)*(1:n)

[Leor Greenberger]

Another question!

I want to do now (assuming an MX8 matrix) column by column addition with vector k transposed:

N=8

k = [0 5 5 10 10 15 15 20]

for r = 1:N

b = k' + a(:,r);

end

Is a for loop the best way?

[Fangjun Jiang]

You just need this:

b=k'+sum(a,2)

[Leor Greenberger]

Oops. I made a mistake in phrasing my question

N=8

k = [0 5 5 10 10 15 15 20]

for r = 1:N

b(:,r) = k' + a(:,r);

end

I want to add each column vector in a with k' and store them in b.

In other words, add column 1 in a with k' and store in column 1 of b, add column 2 in a with k' and store it in column 2 of b, etc.

[the cyclist]

I suggest that accept one of the answers here, assuming that it helped you. And make this comment into a separate question.

[Fangjun Jiang]

For that, you can use repmat.

b=repmat(k',1,N)+a

Answer 2

[the cyclist]

One of many ways to get your result:

M = 7;
N = 5;

V = (-1).^(0:M);

A = bsxfun(@times,1:N,V')

[Andrei Bobrov]

(-1).^(0:7)'*(1:5)

[Jan]

The power operation is very expensive. Using MOD is much faster.

[Andrei Bobrov]

Hi Jan! My "research"

>> t = zeros(100,2);

for j1 = 1:100

tic,(-1).^(0:1000)'*(1:100);t(j1,1)=toc;

tic,(2*rem((1:1000)',2)-1)*(1:100);t(j1,2)=toc;

end

>> [min(t);mean(t);median(t);max(t)]

ans =

0.0008 0.0006

0.0012 0.0012

0.0012 0.0010

0.0030 0.0259

Answer 3

[Sean de Wolski]

b = bsxfun(@plus,k',a(:,1:N))

to your comment in the Fangjun's answer.

Answer 4

[Jonathan]

is there a generic way of making an array of ones that alternate form +1 to -1?

[the cyclist]

Why did you bury a brand-new question as a comment on a 3-year-old thread?