How to do this efficiently?

Hello all,

I have this variable a[k,m]=max(k*Ts+taup,m*Ts+tauq) for k,m=0,1,...,N-1. I want to find the matrix A where [A]_{k,m}=a[k,m] efficiently. To do so, I define two matrices RowInc and ColInc as

RowInc =(0:N-1)'*ones(1,N);
ColInc =transpose(RowInc);

Then I write

for pp=1:Np
    for qq=1:Np
        A=max(RowInc*Ts+tau(pp),ColInc*Ts+tau(qq));
    end
end

Does this give me what I want?

Thanks

[Image Analyst]

Please explain in words what this non-standard MATLAB syntax is supposed to do: [A]_{k,m}=a[k,m]

First of all, MATLAB uses parentheses, not brackets for indexing. It uses brackets but not in the way you used them. Also the underline is not right, nor are your braces and brackets on the left hand side. I have no idea what you intend.

[S. David]

Thanks for your comment. Only the things in gray (using the code environment) are written in MATLAB syntax. Otherwise, it is just mathematical equations written basically using the LateX syntax. So, [A]_{k,m}=a[k,m] means that the (k,m)th element of a matrix A is a[k,m].

[Image Analyst]

Nothing is in gray. But basically you mean that A=a everywhere. The A and a matrices are identical.

[S. David]

When you shade a part of your writing and then press "{}code" in the toolbar the shaded writing will appear in gray, right?

The (k,m)th element of A is a function of k and m for k,m=0,1,...,N-1.

Answer 1

[dpb]

...I want to find the matrix A where [A]{k,m}=a[k,m]..._

A=max(RowInc*Ts+tau(pp),ColInc*Ts+tau(qq));

will end up w/ just a single value for A at the last loop of pp and qq since it overwrites the previous A each iteration. But, I don't believe it does what you want, anyway.

Should be simply

A(A==a);

if I understand the query correctly.

[S. David]

Yes, right. It is a part of larger code. So, the code that corresponds to the above is as following:

RowInc =(0:N-1)'*ones(1,N);
ColInc =transpose(RowInc);

A=max(RowInc*Ts+taup,ColInc*Ts+tauq);

[dpb]

Again, not sure what the question is (or if there even is one)?

Does, or does not, the a array already exist? If so, why is not A result

A(A==a);

as above? Unless you're after a full MxN array only partly populated but if so you've not specified what the remaining elements should be. Presuming zeros,

A=a;
A(A~=a)=0;

[Image Analyst]

His latest comment just said that all elements of A equal "a", so

A(A==a);

net, does absolutely nothing.

[S. David]

Actually, a is not an array, and I need to compute it. I could have written this:

for kk=0:N-1
    for mm=0:N-1
        if kk*Ts+taup>mm*Ts+tauq
           A(kk+1,mm+1)=kk*Ts+taup;
        else
           A(kk+1,mm+1)=mm*Ts+tauq;
        end
     end
end

But N is a large number, e.g. N=2048 and in for loops this takes a while. So, I need to avoid this.

[dpb]

IA, I couldn't figure out what his last comment said (and little of the rest) if after "I have this variable a[k,m]..." there isn't an array a.

Guess I'll leave the field bloodied on this one...

Answer 2

[dpb]

OK, from the loop solution one can write

[x,y]=ndgrid(0:N-1,0:N-1);
A=max(Ts.*x+taup,Ts.*y+tauq);

trading memory for the loop. The loop solution could be simplified since there's only a dependence upon kk for the one term and mm for the other, they could be precomputed outside the loops. Preallocating also would help, of course. Not sure how the timings would come out in the end.

[S. David]

This is exactly what I did, but I wrote

x=(0:N-1)'*ones(1,N);
y=transpose(RowInc);

instead of

[x,y]=ndgrid(0:N-1,0:N-1);

[dpb]

So then what was the question???