How to do this efficiently?

2014 8 9
2 个回答
更新时间：2014 8 11
3 次查看（30 天）

0 个投票

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

4 个评论
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Image Analyst
Image Analyst 2014-8-9
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
S. David 2014-8-9
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
Image Analyst 2014-8-9
Nothing is in gray. But basically you mean that A=a everywhere. The A and a matrices are identical.
S. David
S. David 2014-8-9
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.

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回答（2 个）

dpb
dpb 2014-8-9

1 个投票

...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.

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S. David
S. David 2014-8-9
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
dpb 2014-8-9
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
Image Analyst 2014-8-9
His latest comment just said that all elements of A equal "a", so
A(A==a);
net, does absolutely nothing.
S. David
S. David 2014-8-9
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
dpb 2014-8-9
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...

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dpb
dpb 2014-8-10

0 个投票

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.

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S. David
S. David 2014-8-11
编辑：S. David 2014-8-11
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
dpb 2014-8-11
So then what was the question???

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S. David
S. David
2014-8-9

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dpb
dpb
2014-8-11

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